GaussCombinedContest-Sol.pdf

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2023 Gauss Contests (Grades 7 and 8) Wednesday, May 17, 2023 (in North America and South America) Thursday, May 18, 2023 (outside of North America and South America) Solutions ©2022 University of Waterloo 2023 Gauss Contest Solutions Page 2 2023 Gauss Contest Solutions Page 3 Grade 7 1. Half of 24 is 24 ÷ 2 = 12. Kiyana gives her friend 12 grapes. Answer: (D) 2. Reading from the graph, Friday had the highest temperature. Answer: (C) 3. At a cost of $16.50 a basket, the cost to buy 4 baskets of strawberries is 4 × $16.50 = $66.00. Answer: (B) 4. The difference between 3 and −5 is 3 − (−5) = 3 + 5 = 8. Therefore, it is now 8◦ C warmer. Answer: (A) 5. Since 5 × 5 = 25 and each of the given answers is greater than 25, then the integer that Sarah multiplied by itself must have been greater than 5. Further, 6 × 6 = 36 and each of the given answers is less than or equal to 36. Thus, of the given answers, only 36 could be the result of multiplying an integer by itself. Alternatively, we may have noted that the result of multiplying an integer by itself is a perfect square, and of the answers given, 36 is the only perfect square. Answer: (E) 6. Since the perimeter of P QRS is 40 cm and SR = 16 cm, then the combined length of the remaining three sides is 40 cm − 16 cm = 24 cm. 24 cm = 8 cm. Each of the remaining three sides is equal in length, and so P Q = 3 Answer: (C) 7. Dividing 52 by each of the given denominators, we get that 52 = 13 is the only whole number. 4 Answer: (C) 8. The line segment with greatest length that joins two points on a circle is a diameter of the circle. Since the circle has a radius of 4 cm, then its diameter has length 2 × 4 cm = 8 cm, and so the greatest possible length of the line segment is 8 cm. Answer: (B) 9. The number of integers in the list is 10. Of these integers, 5 are even (they are 10, 12, 14, 16, 5 and 18). Thus, the probability that the chosen integer is even is. 10 Answer: (C) 10. Before adding tax, the combined cost of the three items is $4.20 + $7.60 + $3.20 = $15.00. The 5% tax on $15.00 is 0.05 × $15.00 = $0.75, and so the total cost of the three items, after tax is added, is $15.00 + $0.75 = $15.75. Note that we could have calculated the 5% tax on each individual item, however doing so is less efficient than calculating tax on the $15.00 total. Answer: (D) 2023 Gauss Contest Solutions Page 4 11. Since BCD is a straight line segment, then ∠BCD = 180◦. Therefore, ∠ACB = ∠BCD − ∠ACD = 180◦ − 75◦ = 105◦. Since the sum of the three angles in 4ABC is 180◦ , then ∠ABC = 180◦ − 105◦ − 35◦ = 40◦. Answer: (B) 12. Of the 100 small identical squares, 28 are presently unshaded, and so 100 − 28 = 72 are shaded. So that 75% of the area of W XY Z is shaded, 75 of the 100 small squares must be shaded. Therefore, 75 − 72 = 3 more of the small squares must be shaded. Answer: (A) 13. Suppose we call the unknown vertex V. The side of the rectangle joining the points (2, 1) and (2, 5) is vertical, and so the opposite side of the rectangle (the side joining (4, 1) to V ) must also be vertical. This means that V has the same x-coordinate as (4, 1), which is 4. Similarly, the side of the rectangle joining the points (2, 1) and (4, 1) is horizontal, and so the opposite side of the rectangle (the side joining (2, 5) to V ) must also be horizontal. This means that V has the same y-coordinate as (2, 5), which is 5. Therefore, the coordinates of the fourth vertex of the rectangle are (4, 5). Answer: (D) 14. The prime numbers that are less than 10 are 2, 3, 5, and 7. Thus, the only two different prime numbers whose sum is 10 are 3 and 7. The product of these two numbers is 3 × 7 = 21. Answer: (D) 15. The given list, 2, 9, 4, n, 2n contains 5 numbers. The average of these 5 numbers is 6, and so the sum of the 5 numbers is 5 × 6 = 30. That is, 2 + 9 + 4 + n + 2n = 30 or 15 + 3n = 30, and so 3n = 15 or n = 5. Answer: (D) 16. The sum of P and Q is equal to 5, and so P and Q are (in some order) either equal to 1 and 4, or they are equal to 2 and 3. The difference between R and S is equal to 5, and so R and S must be (in some order) equal to 1 and 6. Since R and S are equal to 1 and 6, then neither P nor Q can be 1, which means that P and Q cannot be equal to 1 and 4, and so they must be equal to 2 and 3. The only numbers from 1 to 6 not accounted for are 4 and 5. Since T is greater than U , then the number that replaces T is 5. Answer: (E) 17. Solution 1 The area of 4AED is equal to one-half its base times its height. Suppose the base of 4AED is AE, then its height is BD (AE is perpendicular to BD). Since AB = BC = 24 cm and E and D are the midpoints of their respective sides, then AE = 12 cm and BD = 12 cm. Thus, the area of 4AED is 21 × 12 cm × 12 cm = 72 cm2. Solution 2 The area of 4AED is equal to the area of 4ABD minus the area of 4EBD. Suppose the base of 4EBD is BD, then its height is EB. Since AB = BC = 24 cm and E and D are the midpoints of their respective sides, then EB = 12 cm and BD = 12 cm. 2023 Gauss Contest Solutions Page 5 Thus, the area of 4EBD is 21 × 12 cm × 12 cm = 72 cm2. The area of 4ABD is equal to 21 × BD × AB = 12 × 12 cm × 24 cm = 144 cm2. Thus, the area of 4AED is 144 cm2 − 72 cm2 = 72 cm2. Answer: (C) 18. The water is in the shape of a rectangular prism with a 2 cm by 5 cm base and depth 6 cm. Therefore, the volume of water is 2 cm × 5 cm × 6 cm = 60 cm3. A face of the prism having the greatest area has dimensions 5 cm by 8 cm. When the prism is tipped so that it stands on a 5 cm by 8 cm face, the water is once again in the shape of a rectangular prism with a 5 cm by 8 cm base and unknown depth. Suppose that after the prism is tipped, the water’s depth is d cm. Since the volume of water is still 60 cm3 when the prism is tipped, then 3 60 =. 5 cm × 8 cm × d cm = 60 cm3 or 40d cm3 = 60 cm3 , and so d = 40 2 When the prism is tipped so that it stands on a face with the greatest area, the depth of the 3 water is cm = 1.5 cm. 2 Answer: (D) 19. Solution 1 We begin by completing a table in which the ones digit of each possible product is listed. For example, when the number on the first die is 3 and the number on the second die is 6, the entry in the table is 8 since 3 × 6 = 18 and the ones digit of 18 is 8. Number on the First Die Number on the Second Die × 1 2 3 4 5 6 1 1 2 3 4 5 6 2 2 4 6 8 0 2 3 3 6 9 2 5 8 4 4 8 2 6 0 4 5 5 0 5 0 5 0 6 6 2 8 4 0 6 Of the 36 possible outcomes in the table above, 6 outcomes have a ones digit that is equal to 0. 6 1 Thus, the probability that the ones digit of the product is 0 is =. 36 6 Solution 2 Since the ones digit of the product is 0, then the product is divisible by 5 and is even. Since the possible numbers in the product are 1, 2, 3, 4, 5, 6, then one of the numbers rolled must be 5. Since the product is even (and 5 is not), then the other number rolled must be one of the three even numbers, namely 2, 4, 6. Thus, the possible pairs of numbers that can be rolled to give a product whose ones digit is 0, 2023 Gauss Contest Solutions Page 6 are (5, 2), (5, 4), (5, 6) or (2, 5), (4, 5), (6, 5). (We note that the first number in the ordered pair represents the first number rolled, while the second number in the pair is the second number rolled.) Since there are 6 possible rolls for each of the two dice, then there are 6 × 6 = 36 possible ordered pairs representing all possible outcomes. Since 6 of these ordered pairs represent a product whose ones digit is 0, then the required 1 6 =. probability is 36 6 Answer: (D) a 2 20. Since a and b are positive integers, then each of and is greater than 0. 7 b a 2 The sum of and is equal to 1, and so each is less than 1. 7 b a Since is greater than 0 and less than 1, then the possible values of a are 1, 2, 3, 4, 5, 6. 7 By substituting each of these values for a into the equation one at a time, we can determine if there is a positive integer value of b for which the equation is true. 1 2 2 1 2 6 Substituting a = 1, we get + = 1 or = 1 − , and so =. 7 b b 7 b 7 2 6 Since = , we can multiply the numerator and denominator of the first fraction by 3 (which b 7 6 6 is 6 ÷ 2) to get =. 3b 7 7 This gives 3b = 7 which does not have an integer solution (b = ). 3 Thus when a = 1, there is no positive integer value of b that satisfies the equation. Substituting a = 2, we get 2 2 2 5 2 2 + = 1 or = 1 − , and so =. 7 b b 7 b 7 2 5 = , we can multiply the numerator and denominator of the first fraction by 5, and b 7 10 10 the numerator and denominator of the second fraction by 2 to get =. 5b 14 This gives 5b = 14 which does not have an integer solution. Thus when a = 2, there is no positive integer value of b that satisfies the equation. Since Substituting a = 3, we get 2 3 2 4 3 2 + = 1 or = 1 − , and so =. 7 b b 7 b 7 2 4 = , we can multiply the numerator and denominator of the first fraction by 2 to get b 7 4 4 =. 2b 7 This gives 2b = 7 which does not have an integer solution. Thus when a = 3, there is no positive integer value of b that satisfies the equation. Since Substituting a = 4 and simplifying, we get 3 2 =. b 7 2 3 = , we can multiply the numerator and denominator of the first fraction by 3, and b 7 6 6 the numerator and denominator of the second fraction by 2 to get =. 3b 14 This gives 3b = 14 which does not have an integer solution. Since 2023 Gauss Contest Solutions Page 7 Thus when a = 4, there is no positive integer value of b that satisfies the equation. 2 2 =. b 7 Since the numerators are equal, then the denominators must be equal, and so b = 7 satisfies the equation. 2 1 Finally, substituting a = 6 and simplifying, we get =. b 7 2 1 Since = , we can multiply the numerator and denominator of the second fraction by 2 to b 7 2 2 get = , and so b = 14. b 14 Thus, there are two pairs of positive integers a and b that satisfy the given equation: a = 5, b = 7 and a = 6, b = 14. Answer: (E) Substituting a = 5 and simplifying, we get 21. Since ABCD is a square and its side lengths are integers, then its area is equal to a perfect square. Since the product of the areas of ABCD and EF GH (the rectangle) is equal to 98, then the area of ABCD is a divisor of 98. The positive divisors of 98 are 1, 2, 7, 14, 49, and 98. There are exactly two divisors of 98 that are perfect squares, namely 1 and 49. Since the area of ABCD is greater than the area of EF GH, then the area of ABCD is 49, and so the area of EF GH is 2 (since 49 × 2 = 98). Square ABCD has area 49, and so AB = BC = CD = DA = 7. The perimeter of ABCDEF GH is equal to = = = = = A G H F E D 7 B 7 7 AB + BC + CD + DE + EF + F G + GH + HA 7 + 7 + 7 + DE + EF + EH + GH + HA (since EH = F G) 21 + DE + EH + HA + EF + GH (reorganizing) 21 + DA + EF + GH (since DE + EH + HA = DA) 21 + 7 + EF + GH (since DA = 7) 28 + 2 × GH (since EF = GH) Since the side lengths are integers and the area of EF GH is 2, then either GH = 1 (and F G = 2), or GH = 2 (and F G = 1). If GH = 1, then the perimeter of ABCDEF GH is 28 + 2 × 1 = 30. Since 30 is not given as a possible answer, then GH = 2 and the perimeter is 28 + 2 × 2 = 32. Answer: (B) 22. If a Gareth sequence begins 10, 8, then the 3rd number in the sequence is 10 − 8 = 2, the 4th is 8 − 2 = 6, the 5th is 6 − 2 = 4, the 6th is 6 − 4 = 2, the 7th is 4 − 2 = 2, the 8th is 2 − 2 = 0, the 9th is 2 − 0 = 2, the 10th is 2 − 0 = 2, and the 11th is 2 − 2 = 0. Thus, the resulting sequence is 10, 8, 2, 6, 4, 2, 2, 0, 2, 2, 0,.... The first 5 numbers in the sequence are 10, 8, 2, 6, 4, the next 3 numbers are 2, 2, 0, and this block of 3 numbers appears to repeat. Since each new number added to the end of this sequence is determined by the two previous numbers in the sequence, then this block of 3 numbers will indeed continue to repeat. (That C 2023 Gauss Contest Solutions Page 8 is, since the block repeats once, then it will continue repeating.) The first 30 numbers of the sequence begins with the first 5 numbers, followed by 8 blocks of 2, 2, 0, followed by one additional 2 (since 5 + 8 × 3 + 1 = 30). The sum of the first 5 numbers is 10 + 8 + 2 + 6 + 4 = 30. The sum of each repeating block is 2 + 2 + 0 = 4, and so the sum of 8 such blocks is 8 × 4 = 32. Thus, the sum of the first 30 numbers in the sequence is 30 + 32 + 2 = 64. Answer: (E) 23. Suppose that the length, or the width, or the height of the rectangular prism is equal to 5. The product of 5 with any of the remaining digits has a units (ones) digit that is equal to 5 or it is equal to 0. This means that if the length, or the width, or the height of the rectangular prism is equal to 5, then at least one of the two-digit integers (the area of a face) has a units digit that is equal to 5 or 0. However, 0 is not a digit that can be used, and each digit from 1 to 9 is used exactly once (that is, 5 cannot be used twice), and so it is not possible for one of the dimensions of the rectangular prism to equal 5. Thus, the digit 5 occurs in one of the two-digit integers (the area of a face). The digit 5 cannot be the units digit of the area of a face, since this would require that one of the dimensions be 5. Therefore, one of the areas of a face has a tens digit that is equal to 5. The two-digit integers with tens digit 5 that are equal to the product of two different one-digit integers (not equal to 5) are 54 = 6 × 9 and 56 = 7 × 8. Suppose that two of the dimensions of the prism are 7 and 8, and so one of the areas is 56. In this case, the digits 5, 6, 7, and 8 have been used, and so the digits 1, 2, 3, 4, and 9 remain. Which of these digits is equal to the remaining dimension of the prism? It cannot be 1 since the product of 1 and 7 does not give a two-digit area, nor does the product of 1 and 8. It cannot be 2 since the product of 2 and 8 is 16 and the digit 6 has already been used. It cannot be 3 since 3 × 7 = 21 and 3 × 8 = 24, and so the areas of two faces share the digit 2. It cannot be 4 since 4 × 7 = 28 and the digit 8 has already been used. Finally, it cannot be 9 since 9 × 7 = 63 and the digit 6 has already been used. Therefore, it is not possible for 7 and 8 to be the dimensions of the prism, and thus 6 and 9 must be two of the three dimensions. Using a similar systematic check of the remaining digits, we determine that 3 is the third dimension of the prism. That is, when the dimensions of the prism are 3, 6 and 9, the areas of the faces are 3 × 6 = 18, 3 × 9 = 27, and 6 × 9 = 54, and we may confirm that each of the digits from 1 to 9 has been used exactly once. Since the areas of the faces are 18, 27 and 54, the surface area of the rectangular prism is 2 × (18 + 27 + 54) or 2 × 99 = 198. Answer: (D) 2023 Gauss Contest Solutions Page 9 24. Solution 1 Begin by colouring the section at the top blue. Since two circles have the same colouring if one can be rotated to match the other, it does not matter which section is coloured blue, so we arbitrarily choose the top section. blue There are now 5 sections which can be coloured green. After choosing the section to be coloured green, there are 4 sections remaining which can be coloured yellow. Each of the remaining 3 sections must then be coloured red. Thus, the total number of different colourings of the circle is 5 × 4 = 20. Solution 2 We begin by considering the locations of the three sections coloured red, relative to one another. The three red sections could be adjacent to one another, or exactly two red sections could be adjacent to one another, or no red section could be adjacent to another red section. We consider each of these 3 cases separately. Case 1: All three red sections are adjacent to one another Begin by colouring any three adjacent sections red. red red red Since two circles have the same colouring if one can be rotated to match the other, it does not matter which three adjacent sections are coloured red. Consider the first section that follows the three red sections as we move clockwise around the circle. There are 3 choices for the colour of this section: blue, green or yellow. Continuing to move clockwise to the next section, there are now 2 choices for the colour of this section. Finally, there is 1 choice for the colour of the final section, and thus there are 3 × 2 × 1 = 6 different colourings of the circle in which all three red sections are adjacent to one another. These 6 colourings are shown below. red red blue yellow red red red green yellow blue red red green blue red red red yellow green green red blue red red yellow green red yellow red blue red red red yellow green blue 2023 Gauss Contest Solutions Page 10 Case 2: Exactly two red sections are adjacent to one another There are two different possible arrangements in which exactly two red sections are adjacent to one another. In the first of these, the next two sections that follow the two adjacent red sections as we move clockwise around the circle, are both not red. We call this Case 2a. In the second of these, the section that follows the two adjacent red sections as we move clockwise around the circle is not red, but the next section is. We call this Case 2b. The arrangements for Cases 2a and 2b are shown below. red red red red red red Case 2a Case 2b Notice that the first of these two circles cannot be rotated to match the second. The number of colourings in Case 2a and in Case 2b are each equal to the number of colourings as in Case 1. That is, there are 3 choices for the first uncoloured section that follows the two red sections as we move clockwise around the circle. Continuing to move clockwise to the next uncoloured section, there are now 2 choices for the colour of this section. Finally, there is 1 choice for the colour of the final section, and thus there are 3 × 2 × 1 = 6 different colourings of the circle in Case 2a as well as in Case 2b. These 12 colourings are shown below. red Case 2a red blue red green red yellow yellow red red Case 2b red red red red yellow yellow red blue blue green red red blue red green blue green blue red yellow green red red yellow green red red red green yellow red red green blue red yellow red blue red red green red blue red yellow yellow blue red red red green yellow green red red blue red red yellow blue green red Case 3: No red section is adjacent to another red section Begin by colouring any three non-adjacent sections red. red red red Since two circles have the same colouring if one can be rotated to match the other, it does not matter which three non-adjacent sections are coloured red. In this case, there are 2 possible colourings as shown below. 2023 Gauss Contest Solutions Page 11 green red blue red yellow red red yellow blue red green red A circle with any other arrangement of the green, yellow and blue sections can be rotated to match one of the two circles above. The total number of different colourings of the circle is 6 + 12 + 2 = 20. Answer: (E) 25. We can represent the given information in a Venn diagram by first introducing some variables. Let x be the number of students that participated in hiking and canoeing, but not swimming. Let y be the number of students that participated in hiking and swimming, but not canoeing. Let z be the number of students that participated in canoeing and swimming, but not hiking. Since 10 students participated in all three activities and no students participated in fewer than two activities, we complete the Venn diagram as shown. Total: n participants swimming hiking y 0 x 10 0 z 0 canoeing Suppose that the total number of students participating in the school trip was n. 50 n Since 50% of all students participated in at least hiking and canoeing, then n or partici100 2 pated in at least hiking and canoeing. n Since this number of students, , is an integer, then n must be divisible by 2. 2 60 3n Similarly, n or students participated in at least hiking and swimming. 100 5 3n Since this number of students, , is an integer, then n must be divisible by 5 (since 3 and 5 5 have no factors in common). This means that n is divisible by both 2 and 5, and thus n is divisible by 10. n 3n From the Venn diagram, we see that x + 10 = , and y + 10 =. 2 5 Since the total number of participants is n, we also get that x+y+z+10 = n or z = n−10−x−y. We may now use these equations, x= n 3n − 10, y = − 10, and z = n − 10 − x − y 2 5 and the fact that n is divisible by 10, to determine all possible values of z. We can then use the values of z to determine all possible values of the positive integer k, where k% participated in at least canoeing and swimming. Since n is a positive integer that is divisible by 10, its smallest possible value is 10. n However, substituting n = 10 into x = − 10, we get x = 5 − 10 and so x = −5 which is not 2 possible. (Recall that x is the number of students that participated in hiking and canoeing, but not swimming, and so x ≥ 0.) Next, we try n = 20. 2023 Gauss Contest Solutions Page 12 When n = 20, x = 10 − 10 and so x = 0. 3 × 20 When n = 20, y = − 10 or y = 12 − 10, and so y = 2. 5 Finally, when n = 20, x = 0, and y = 2, we get z = 20 − 10 − 0 − 2 = 8. When z = 8, the number of students who participated in at least canoeing and swimming is 8 + 10 = 18 (since 10 students participated in all three), and so the percentage of students who 18 × 100% = 90%, and so k = 90. participated in at least canoeing and swimming is 20 In the table below, we continue in this way by using successively greater multiples of 10 for the value of n. n x= n − 10 2 y= 3n − 10 5 z = n − 10 − x − y k= k= z + 10 × 100 n 8 + 10 × 100 = 90 20 20 0 2 8 30 5 8 7 40 10 14 6 k= 6 + 10 × 100 = 40 40 50 15 20 5 k= 5 + 10 × 100 = 30 50 60 20 26 4 k= 4 + 10 × 100 ≈ 23.3 60 70 25 32 3 k= 3 + 10 × 100 ≈ 18.6 70 80 30 38 2 90 35 44 1 100 40 50 0 k= 7 + 10 × 100 ≈ 56.7 30 k= k= 2 + 10 × 100 = 15 80 1 + 10 × 100 ≈ 12.2 90 k= 0 + 10 × 100 = 10 100 For values of n that are greater than 100, we get that z < 0, which is not possible. Therefore, the sum of all such positive integers k is 90 + 40 + 30 + 15 + 10 = 185. Answer: (B) 2023 Gauss Contest Solutions Page 13 Grade 8 1. The fraction 1 4 is equivalent to 1 ÷ 4 = 0.25. Answer: (B) 2. Reading from the graph, the forecast wind speed is less than 20 km/h on Monday, Tuesday, Wednesday, and Sunday. Thus, Jack will be able to sail alone on 4 days during this 7-day period. Answer: (A) 3. We note that 15 × 10 = 150, 15 × 2 = 30, 15 × 3 = 45, and 15 × 4 = 60. Since there is no integer n for which 15 × n = 25, then 25 is not a multiple of 15. Answer: (B) 4. Ordering the given list of integers from least to greatest, we get −9, −7, 0, 9, 10. The third integer in the ordered list is 0. Answer: (D) 5. Solution 1 14 = 7. If 2n = 14, then n = 2 When n = 7, the value of 10n is 10 × 7 = 70. Solution 2 Multiplying both sides of the given equation by 5, we get 5 × 2n = 5 × 14, and so 10n = 70. Answer: (C) 6. There are 6 possible outcomes when Tallulah rolls a single standard die once. 1 2 =. 6 3 Answer: (A) She loses if she rolls 2 of these 6 outcomes, and so the probability that she loses is 7. Solution 1 We may convert the given addition problem to a subtraction problem. That is, since 1013 + P QP Q = 2023, then P QP Q = 2023 − 1013. The difference between 2023 and 1013 is 2023 − 1013 = 1010, and so P = 1, Q = 0, and P + Q = 1 + 0 = 1. Solution 2 The ones (units) digit of the sum 2023 is 3. Thus, the ones digit of the sum 3 + Q must equal 3. Since Q is a digit, the only possible value of Q is 0. The tens digit of the sum 2023 is 2. Since there is no “carry” from the ones column to the tens column, the ones digit of the sum 1 + P must equal 2. Since P is a digit, the only possible value of P is 1. We may confirm that when P = 1 and Q = 0, we get 1013 + 1010 = 2023 as required. The value of P + Q is 1 + 0 = 1. Answer: (B) 2023 Gauss Contest Solutions Page 14 8. Suppose the salad dressing initially contains 300 mL of oil. Since the ratio of oil to vinegar is 3 : 1, then the salad dressing initially contains one-third as 1 much vinegar as oil, or × 300 mL = 100 mL of vinegar (note that 300 : 100 = 3 : 1). 3 If the volume of vinegar is doubled, the new volume of vinegar is 2 × 100 mL = 200 mL, and so the new ratio of oil to vinegar is 300 : 200 = 3 : 2. Note: We chose to begin with 300 mL of oil, but performing the above calculations with any starting volume of oil will give the same 3 : 2 ratio. Answer: (A) 9. Before including tax, the combined cost of the three items is $4.20 + $7.60 + $3.20 = $15.00. The 5% tax on $15.00 is 0.05×$15.00 = $0.75, and so the total cost of the three items, including tax, is $15.00 + $0.75 = $15.75. Note that we could have calculated the 5% tax on each individual item, however doing so is less efficient than calculating tax on the $15.00 total. Answer: (D) 10. When (1, 3) is reflected in the y-axis, the reflected point is (−1, 3). In general, when a point is reflected in the y-axis, its x-coordinate changes sign, and its y-coordinate does not change. Thus, the vertices of the reflected rectangle are (−1, 3), (−1, 1), (−4, 1), and (−4, 3). Of the given possibilities, (−3, 4) is not a vertex of the reflected rectangle. Answer: (C) 11. In the leftmost rectangle, the length of the path along the rectangle’s diagonal, d, and the sides with lengths 3 and 4, form a right-angled triangle. √ √ Using the Pythagorean Theorem, we get d2 = 32 + 42 , and so d = 32 + 42 = 25 = 5 (these are “3-4-5” right-angled triangles). The path from A to B includes one such diagonal, two vertical sides each of length 4, and three horizontal sides each of length 3, and thus has length 5 + (2 × 4) + (3 × 3) = 22. Answer: (A) 12. Since ∠P QR is a straight angle, its measure is 180◦ , and so ∠SQR = 180◦ − 125◦ = 55◦. Since SQ = SR, then ∠SRQ = ∠SQR = 55◦. The sum of the angles inside 4SQR is 180◦ , and so x = 180 − 55 − 55 = 70. Answer: (B) 13. The number of peaches in the original pile is two more than a multiple of three. Of the choices given, 29 is the only number which is two more than a multiple of three (29 = 3 × 9 + 2). Answer: (D) 14. The sum of each block of 5 repeating integers is 4 − 3 + 2 − 1 + 0 = 2. In the first 23 integers, there are 4 such blocks of 5 integers, plus 3 additional integers (since 23 = 4 × 5 + 3). The sum of the 4 blocks of 5 repeating integers is 4 × 2 = 8, and the next three integers in the list following these 4 blocks are 4, −3, 2. Thus, the sum of the first 23 integers is 8 + 4 − 3 + 2 = 11. Answer: (D) 2023 Gauss Contest Solutions Page 15 15. The circumference of each of Bindu’s bike tires is 2 × π × 30 cm = 60π cm. If the bike tires rotate exactly five times, then the distance travelled by Bindu’s bike is 5 × 60π cm = 300π cm. Answer: (D) 16. Solution 1 The sum of all 8 numbers in the list is 41 + 35 + 19 + 9 + 26 + 45 + 13 + 28 = 216. When the 8 numbers are arranged in pairs, there are 4 pairs. 216 = 54. The sum of the numbers in each pair is the same, and so this sum is 4 Therefore, the number paired with 13 is 54 − 13 = 41. Note: We may confirm that 45 + 9 = 54, 41 + 13 = 54, 35 + 19 = 54, and 28 + 26 = 54. Solution 2 Since the sum of the numbers in each pair is the same, then the largest number in the list must be paired with the smallest number, the second largest with the second smallest, and so on. (Can you reason why this must be true?) That is, the largest and smallest numbers in the list, 45 and 9, must be paired. The second largest and second smallest numbers, 41 and 13, must be paired, and so the number paired with 13 is 41. Note: We may confirm that 45 + 9 = 54, 41 + 13 = 54, 35 + 19 = 54, and 28 + 26 = 54. Answer: (E) 17. The mean (average) is determined by adding the 30 recorded temperatures, and dividing the sum by 30. The sum of the temperatures for the first 25 days was 25 × 21◦ C = 525◦ C. The sum of the temperatures for the last 5 days was 5 × 15◦ C = 75◦ C. 600◦ C 525◦ C + 75◦ C = = 20◦ C. Thus, the mean of the recorded temperatures was 30 30 Answer: (C) 18. We begin by listing, in order, the smallest 2-digit positive divisors of 630. We note that by first writing 630 as a product of its prime factors (630 = 2 × 32 × 5 × 7), it may be easier to determine these divisors. The smallest five 2-digit positive divisors of 630 are 10, 14, 15, 18, and 21. The next largest 2-digit positive divisor of 630 is 30, and we notice that 21 × 30 = 630. That is, 21 and 30 are a pair of 2-digit positive integers whose product is 630, and they are consecutive in the ordered list of positive divisors. Thus, each of 10, 14, 15, and 18 must be paired with a divisor of 630 that is greater than 30. We may check that the divisors that pair with each of 10, 14, 15, and 18 is a 2-digit positive integer by dividing 630 by the smaller divisor. 630 630 630 630 = 35, = 42, = 45, and = 63. That is, 18 15 14 10 Thus, the pairs of 2-digit positive integers whose product is 630 are 21 and 30, 18 and 35, 15 and 42, 14 and 45, and 10 and 63, and so there are 5 such pairs. Answer: (D) 2023 Gauss Contest Solutions Page 16 7 4 3 7 1 19. Between 9 a.m. and 10 a.m., Ryan cut − = − = of his lawn. 8 2 8 8 8 3 1 60 Ryan cut of his lawn in 1 hour (60 minutes), and so he cut of his lawn in minutes or 8 8 3 20 minutes. 7 7 1 At 10 a.m., Ryan had cut of his lawn, and thus had 1 − = of his lawn left to cut. 8 8 8 1 Since Ryan cuts of his lawn in 20 minutes, then he finished at 10:20 a.m. 8 Answer: (C) 20. Begin by placing four tiles in the squares of the first row. The only restriction is that the row must contain one tile of each colour. Thus in the first row, there are 4 choices of tile colour for the first column, 3 choices for the second, 2 for the third, and 1 choice for the fourth. That is, there are 4 × 3 × 2 × 1 = 24 different ways to cover the squares in the first row using one tile of each of the four colours. For example, using R for red, B for black, G for green, and Y for yellow, the first row could contain tiles coloured G Y R B, in that order. Suppose that the first row does contain tiles coloured G Y R B, in that order. We will show that there is only one way to arrange the remaining 12 tiles in the grid. Consider the first square in the second row, that is, the square directly below the tile coloured G. The tile placed in this square cannot be coloured G since it shares an edge with the tile coloured G in row 1. Also, the tile placed in this square cannot be coloured Y since it touches the corner of the square containing the tile coloured Y in row 1. Assume that the tile in the first square of row 2 is coloured B, as shown. G Y R B Next, consider the colour of the tiles that could be placed in the B second square of row 2. The tile in this square cannot be coloured Y since it shares an edge with the tile coloured Y in row 1. Also, the tile in this square cannot be coloured G since it touches the corner of the square containing the tile coloured G in row 1. Further, the tile in this square cannot be coloured R since it touches the corner of the square containing the tile coloured R in row 1. Since the first square in this row contains a tile coloured B, then we have no possible tile that can be placed in the second square of row 2. This means that the tile in the first square of row 2 cannot be coloured B, and thus it must be coloured R, as shown. G Y R B The tile in the second square of row 2 cannot be coloured Y or R G (as noted earlier), and thus must be coloured B. Continuing to move right along row 2, the next tile cannot be coloured Y since it touches the corner of the square containing the tile coloured Y in row 1, and so the tile in this square must be coloured G, with the final tile in the row being coloured Y. That is, the positions of the 4 tiles in row 2 are completely determined by the tiles in row 1. Thus for each of the 24 different ways to place the tiles in row 1, there is exactly one way to place the tiles in row 2. Repeating the argument, the same is then true for the tiles in row 3 and row 4; that is, there 2023 Gauss Contest Solutions Page 17 is exactly one choice for the location of each of the coloured tiles within each of these two rows as well. The 4 × 4 in our example above is completed here, as shown. G Y R B You should justify for yourself that each of rows 3 and 4 must R B G Y contain tiles exactly as shown. G Y R B For each of the 24 different ways to cover the squares in the first R B G Y row using one tile of each of the four colours, there is exactly one way to cover all remaining squares in the grid. Thus, there are 24 different ways that the tiles can be arranged. Answer: (B) 21. Since OM is a radius of the circle, then OM = 87. 4M N O is a right-angled triangle, and so by the Pythagorean Theorem, we get OM 2 = M N 2 + N O2 or 872√= 632 + N O2 , and so N O2 = 872 − 632 = 3600. Since N O > 0, then N O = 3600 = 60. Since OP is also a radius, then OP = 87, and so N P = N O + OP = 60 + 87 = 147. The area of 4P M N is equal to 21 × N P × M N = 12 × 147 × 63 = 4630.5. Answer: (D) 22. Nasrin’s mean (average) speed is determined by dividing the total distance travelled, which is 9 km, by the total time. It took Nasrin 2 hours and thirty minutes, or 150 minutes, to canoe into her camp. 1 On the return trip, it took her × 150 minutes or 50 minutes. 3 Thus, the total time for Nasrin to paddle to camp and back was 200 minutes. 20 1 Converting to hours, 200 minutes is 3 hours and 20 minutes, and since 20 minutes is = 60 3 hours, it took Nasrin 3 13 hours in total. 9 km 9 km Thus, Nasrin’s mean speed as she paddled to camp and back was 1 or 10 , which is equal h 33 h 3 3 27 to 9 × km/h = km/h = 2.7 km/h. 10 10 Answer: (E) 23. To begin, the volume of water in Cylinder B is π × (8 cm)2 × 50 cm = 3200π cm3. After some water is poured from Cylinder B into Cylinder A, the total volume of water in the two cylinders will be 3200π cm3 (since no water is lost). Let h cm be the height of the water in each of the two cylinders when the height of the water in both cylinders is the same. At this time, the volume of water in Cylinder B is π × (8 cm)2 × h cm = 64πh cm3. At this time, the volume of water in Cylinder A is π × (6 cm)2 × h cm = 36πh cm3. Thus, the total volume of water in the two cylinders is 64πh cm3 + 36πh cm3 = 100πh cm3 , 3200π and so 100πh = 3200π or h = = 32. 100π When the height of the water in both cylinders is the same, that height is 32 cm. Answer: (C) 2023 Gauss Contest Solutions Page 18 24. Solution 1 a b We begin by multiplying the given equation through by 20 to get 20 × + 20 × = 20 × 7, 4 10 or 5a + 2b = 140. Since 5a = 140 − 2b and both 140 and 2b are even, then 5a is even, which means that a is even. We start by trying a = 20 and b = 20 which is a solution, since 5 × 20 + 2 × 20 = 140. This pair for a and b satisfies all of the conditions except a < b. We may find other solutions to the equation 5a + 2b = 140 by adding two 5s and subtracting five 2s (this is the same as adding 10 and subtracting 10), or by subtracting two 5s and adding five 2s. Adding two 5s is equivalent to increasing the value of a by 2. Subtracting five 2s is equivalent to decreasing the value of b by 5. Consider a = 20 + 2 = 22 and b = 20 − 5 = 15. This is a solution since 5 × 22 + 2 × 15 = 140. However, in this case a > b and every time we add two 5s and subtract five 2s, a becomes greater and b becomes smaller. Thus, we need to go in the other direction. Consider a = 20 − 2 = 18 and b = 20 + 5 = 25. This is a solution since 5 × 18 + 2 × 25 = 140. Here, a < b and a + b = 43, so this pair satisfies all of the conditions. Next, consider a = 18 − 2 = 16 and b = 25 + 5 = 30. This is a solution since 5 × 16 + 2 × 30 = 140. Here, a < b and a + b = 46, so this pair satisfies all of the conditions. Notice that the sum a + b increases by 3 on each of these steps. This means that doing this 17 more times gets us to a = 16 − 17 × 2 = −18 and b = 30 + 17 × 5 = 115. This is a solution since 5 × (−18) + 2 × 115 = 140. Notice that it is still the case that a < b and a + b < 100. Repeating this process one more time, we get a = −20 and b = 120, which gives a + b = 100 and so there are no more pairs that work. Since we know that a has to be even and we are considering all possible even values for a, there can be no other pairs that work. In total, there are 1 + 1 + 17 = 19 pairs of integers a and b that satisfy each of the given conditions and the given equation. Solution 2 We begin by rearranging Doing so, we get b a + 4 10 b 10 b 10 × 10 the given equation to isolate b. =7 =7− a 4 = 10 × 7 − 10 × a 4 (multiplying each term by 10) 10a 4 5a b = 70 − 2 5a 5a Since b is an integer, then 70 − is an integer, which means that must be an integer. 2 2 b = 70 − 2023 Gauss Contest Solutions Page 19 Since 2 does not divide 5, then 2 must divide a and so a is even. 5a Since a < b and b = 70 − , then 2 a < 70 − 5a 2 2 × a < 2 × 70 − 2 × 5a 2 (multiplying each term by 2) 2a < 140 − 5a 7a < 140 a < 20 Further, since a + b < 100 and b = 70 − 5a , then 2 5a < 100 2 5a 2 × a + 2 × 70 − 2 × < 2 × 100 2 2a + 140 − 5a < 200 −60 < 3a −20 < a a + 70 − (multiplying each term by 2) Thus a is an even integer that is greater than −20 and less than 20. Since there are 19 even integers from −18 to 18 inclusive, we suspect that there are 19 pairs of integers a and b that satisfy the given equation. It is a good idea (and good practice) to at least check that the largest and smallest of these values of a do indeed satisfy each of the given conditions. 5(−18) or b = 70 − 5(−9), and so b = 115. When a = −18, we get b = 70 − 2 This pair satisfies the given conditions that a < b and a + b < 100. Substituting a = −18 and b = 115 into the given equation, we get b −18 115 −9 23 14 a + = + = + = = 7, 4 10 4 10 2 2 2 and thus a = −18 and b = 115 is a solution. 5(18) When a = 18, we get b = 70 − or b = 70 − 5(9), and so b = 25. 2 This satisfies the given conditions that a < b and a + b < 100. Substituting a = 18 and b = 25 into the given equation, we get a b 18 25 9 5 14 + = + = + = = 7, 4 10 4 10 2 2 2 and thus a = 18 and b = 25 is also a solution. At this point we can be confident that for each of the 19 even integer values of a from −18 to 18 inclusive, there is an integer b for which the pair of integers a and b satisfy each of the given conditions and the given equation. Answer: (B) 2023 Gauss Contest Solutions Page 20 25. For any triangle, the sum of the lengths of two sides is always greater than the length of the third side. This property is known as the triangle inequality. If for example the side lengths of a triangle are a, b and c, then the triangle inequality says that a + b > c and a + c > b and b + c > a. We begin by considering the number of different ways to choose three integers from 3, 4, 10, 13 (without using n), and then forming a triangle whose side lengths are equal to those integers. Consider choosing the integers 3, 4, 10. Since 3 + 4 < 10, then it is not possible to form a triangle whose side lengths are 3, 4, 10. Consider choosing the integers 3, 4, 13. Since 3 + 4 < 13, then it is not possible to form a triangle whose side lengths are 3, 4, 13. Consider choosing the integers 3, 10, 13. Since 3 + 10 = 13, then it is not possible to form a triangle whose side lengths are 3, 10, 13. However, the remaining possible choice of three side lengths, 4, 10, 13, does satisfy the triangle inequality since 4 + 10 > 13 and 4 + 13 > 10 and 10 + 13 > 4. Thus, without using the value of n, there is exactly one way to choose three integers and form a triangle whose side lengths are equal to those integers. This means that we need to determine values of n for which there are exactly three different ways to choose two of the integers 3, 4, 10, 13 and form a triangle whose side lengths are equal to those two integers and n. There are six possible ways to choose two integers from the list 3, 4, 10, 13. Thus for each value of n, the triangles we need to consider have side lengths: 3, 4, n or 3, 10, n or 3, 13, n or 4, 10, n or 4, 13, n or 10, 13, n. For each value of n, we need exactly three of these six triangles to satisfy the triangle inequality. Next, we use the triangle inequality to determine the restrictions on n for each of the six possible groups of triangles. In a triangle with side lengths 3, 4, n, we get 3 + n > 4 or n > 1, and 3 + 4 > n or n < 7, and 4 + n > 3 or n > −1. To satisfy all three inequalities, n must be greater than 1 and less than 7. Thus, the possible values of n for which a triangle has side lengths 3, 4, n are n = 2, 3, 4, 5, 6. Since n must be different from all other numbers in the list, then n = 2, 5, 6. In a triangle with side lengths 3, 10, n, we get 3 + n > 10 or n > 7, and 3 + 10 > n or n < 13, and 10 + n > 3 or n > −7. To satisfy all three inequalities, n must be greater than 7 and less than 13. Thus, the possible values of n for which a triangle has side lengths 3, 10, n are n = 8, 9, 11, 12 (n 6= 10 since 10 is in the list). In a triangle with side lengths 3, 13, n, we get 3 + n > 13 or n > 10, and 3 + 13 > n or n < 16, and 13 + n > 3 or n > −10. To satisfy all three inequalities, n must be greater than 10 and less than 16. Thus, the possible values of n for which a triangle has side lengths 3, 13, n are n = 11, 12, 14, 15 (n 6= 13 since 13 is in the list). In a triangle with side lengths 4, 10, n, we get 4 + n > 10 or n > 6, and 4 + 10 > n or n < 14, and 10 + n > 4 or n > −6. To satisfy all three inequalities, n must be greater than 6 and less than 14. Thus, the possible values of n for which a triangle has side lengths 4, 10, n are n = 7, 8, 9, 11, 12. In a triangle with side lengths 4, 13, n, we get 4 + n > 13 or n > 9, and 4 + 13 > n or n < 17, and 13 + n > 4 or n > −9. To satisfy all three inequalities, n must be greater than 9 and less than 17. 2023 Gauss Contest Solutions Page 21 Thus, the possible values of n for which a triangle has side lengths 4, 13, n are n = 11, 12, 14, 15, 16. In a triangle with side lengths 10, 13, n, we get 10 + n > 13 or n > 3, and 10 + 13 > n or n < 23, and 13 + n > 10 or n > −3. To satisfy all three inequalities, n must be greater than 3 and less than 23. Thus, the possible values of n for which a triangle has side lengths 10, 13, n are n = 5, 6, 7, 8, 9, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22. Recall that for each value of n, we need exactly three of the six triangles to satisfy the triangle inequality (the triangle with side lengths 4, 10, 13 is the fourth). Clearly for values of n less than 7, there are too few triangles, and similarly for values of n greater than 16, there are also too few triangles. (There are at most two triangles in each of these two cases.) In the table below, we summarize our work by placing a checkmark if the triangle satisfies the triangle inequality and then counting the number of such triangles. n 7 8 9 11 12 14 15 16 (3, 4, n) (3, 10, n) X X X X (3, 13, n) X X X X (4, 10, n) X X X X X (4, 13, n) (10, 13, n) (4, 10, 13) X X X X X X X X X X X X X X X X X X X X X number of triangles 3 4 4 6 6 4 4 3 Therefore, there are exactly four different values of n that satisfy the given conditions, and the sum of these values of n is 8 + 9 + 14 + 15 = 46. Answer: (A) The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2022 Gauss Contests (Grades 7 and 8) Wednesday, May 18, 2022 (in North America and South America) Thursday, May 19, 2022 (outside of North America and South America) Solutions ©2021 University of Waterloo 2022 Gauss Contest Solutions Page 2 Centre for Education in Mathematics and Computing Faculty and Staff Ed Anderson Jeff Anderson Terry Bae Jacquelene Bailey Shane Bauman Ersal Cahit Diana Castañeda Santos Sarah Chan Ashely Congi Serge D’Alessio Fiona Dunbar Mike Eden Sandy Emms Barry Ferguson Steve Furino Lucie Galinon Robert Garbary Rob Gleeson Sandy Graham Conrad Hewitt Lisa Kabesh Jenn Kelebuda Carrie Knoll Wesley Korir Judith Koeller Laura Kreuzer Bev Marshman Josh McDonald Paul McGrath Comfort Mintah Jen Nelson Ian Payne J.P. Pretti Alexandra Rideout Nick Rollick Kim Schnarr Tucker Seabrook Ashley Sorensen Ian VanderBurgh Troy Vasiga Christine Vender Heather Vo Bonnie Yi Gauss Contest Committee Ashley Sorensen (Chair), University of Waterloo, Waterloo, ON Kevin Grady (Assoc. Chair), Cobden, ON Sarah Garrett, Mitchell Woods P.S., Guelph, ON Kora Lee Gallant, Madeline Symonds M.S., Hammonds Plains, NS JoAnne Halpern, Toronto, ON Clay Kellough, Fort Richmond C.I., Winnipeg, MB David Matthews, Waterloo, ON John McLoughlin, University of New Brunswick, Fredericton, NB Nick Rollick, University of Waterloo, Waterloo, ON David Switzer, Scott Central P.S., Sandford, ON Rachael Verbruggen, University of Waterloo, Waterloo, ON Robert Wong, Edmonton, AB Chris Wu, Ledbury Park E. and M.S., Toronto, ON Lori Yee, William Dunbar P.S., Pickering, ON 2022 Gauss Contest Solutions Page 3 Grade 7 1. Arranging the five answers along with 10 from smallest to largest, we get 1, 5, 8, 10, 13, 19. Since 10 is 2 more than 8 and 10 is 3 less than 13, then 8 is the closest number to 10. Answer: (C) 2. Reading from the graph, the greatest number of hours that Gabe spent riding his bike is 4, and this occurred on Tuesday. Answer: (B) 3. Of the given answers, 0 is the only value of x that is less than 5. Answer: (B) 4. Since 18 + 5 = 23, and 23 + 5 = 28, and 28 + 5 = 33, the next three terms in the sequence are 23, 28, 33. Answer: (C) 5. The faces shown are labelled with 1, 3 and 5 dots. Therefore, the other three faces are labelled with 2, 4 and 6 dots. The total number of dots on the other three faces is 2 + 4 + 6 = 12. Answer: (D) 6. Since ∠ABC measures 90◦ , and ∠ABC = 44◦ + x◦ , then x = 90 − 44 = 46. Answer: (A) 7. The largest height of the singers in Saura’s choir is 183.5 cm. The smallest height of the singers in Saura’s choir is 141 cm. Thus, the range of their heights is 183.5 cm − 141 cm = 42.5 cm. Answer: (A) 8. Beginning at the origin (0, 0), the point (3, −4) is located right 3 units and down 4 units. In the diagram, the point (3, −4) is labelled T. Answer: (E) 9. When Emily jumps for 75 seconds, she jumps for 60 + 15 seconds. Jumping at the rate of 52 times in 60 seconds, Emily jumps 52 ÷ 4 = 13 times in 60 ÷ 4 = 15 seconds. Since Emily jumps 52 times in 60 seconds and 13 times in 15 seconds, then Emily jumps 52 + 13 = 65 times in 60 + 15 = 75 seconds. Answer: (C) $1.00 = 10 dimes. $0.10 $1.00 In $1.00 worth of quarters, there are = 4 quarters. $0.25 The jar contains 10 dimes and a total of 10 + 4 = 14 coins. 10. In $1.00 worth of dimes, there are 10 = 14 Answer: If Terry randomly removes one coin from the jar, the probability that it is a dime is 5. 7 (E) 2022 Gauss Contest Solutions Page 4 11. Since 42 is an even number, then 2 is a factor of 42. Since 42 = 2 × 21 and 21 = 3 × 7, then 42 = 2 × 3 × 7. Each of 2, 3 and 7 is a prime number and each is a factor of 42. Thus, the sum of the prime factors of 42 is 2 + 3 + 7 = 12. (We note that 1, 6, 14, 21, and 42 are also factors of 42, however they are not prime numbers.) Answer: (C) 12. 4P QR is isosceles with P Q = P R, and so ∠P RQ = ∠P QR. The sum of the angles in 4P QR is 180◦. Since ∠QP R = 70◦ , then the other two angles in this triangle add to 180◦ − 70◦ = 110◦. Since these two angles are equal, they each measure 110◦ ÷ 2 = 55◦ , and so x = 55. Since QRST is a rectangle, each of its interior angles is a right angle, and so y = 90. The value of x + y is 55 + 90 = 145. Answer: (D) 13. A two-digit number has at least one digit that is a 4 if its tens digit is a 4 or if its ones digit is a 4. There are 10 two-digit numbers whose tens digit is a 4. These are 40, 41, 42, 43, 44, 45, 46, 47, 48, and 49. There are 9 two-digit numbers whose ones digit is a 4. These are 14, 24, 34, 44, 54, 64, 74, 84, 94, including 44 which was counted in our previous list. The number of two-digit numbers that have at least one digit that is a 4 is 10 + 9 − 1 = 18. Answer: (C) 14. The side lengths of each of the three identical squares are equal. The perimeter of W XY Z is 56 m and is comprised of 8 such side lengths. Thus, the length of each side of the three identical squares is 56 m ÷ 8 = 7 m. The area of each of the three identical squares is 7 m × 7 m = 49 m2. Therefore, the area of W XY Z is 3 × 49 m2 = 147 m2. Answer: (B) 15. The first Wednesday of a month must occur on one of the first 7 days of a month, that is, the 1st , 2nd , 3rd , 4th , 5th , 6th , or 7th of the month. The second Wednesday of a month occurs 7 days following the first Wednesday of that month, and the third Wednesday of a month occurs 14 days following the first Wednesday of that month. Adding 14 days to each of the possible dates for the first Wednesday of the month, we get that the third Wednesday of a month must occur on the 15th , 16th , 17th , 18th , 19th , 20th , or 21st of that month. Of the given answers, the public holiday cannot occur on the 22nd of that month. Answer: (B) 16. Solution 1 When a standard fair coin is tossed three times, there are 8 possible outcomes. If H represents head and T represents tail, these 8 outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. Of these 8 outcomes, there are exactly 2 whose outcomes are all the same (HHH and TTT). Therefore, the probability that the three outcomes are all the same is 28 = 14. 2022 Gauss Contest Solutions Page 5 Solution 2 The three outcomes are all the same exactly when each of the three tosses is a head or when each of the three tosses is a tail. When a coin is tossed, the probability that the coin shows a head is 21 and the probability that the coin shows a tail is also 12 (there are two possible outcomes and each is equally probable). Thus, the probability that each of the three tosses is a head is 12 × 21 × 12 = 18. Similarly, the probability that each of the three tosses is a tail is 21 × 12 × 21 = 18. Therefore, the probability that the three outcomes are all the same is 18 + 81 = 28 = 14. Answer: (C) 17. If the value of P is less than 9, then QR + P P P + P P P is at most 99 + 888 + 888 = 1875. Since the given sum is 2022, then the value of P cannot be less than 9 and thus must equal 9. When P = 9, we get QR + 999 + 999 = 2022, and so QR = 2022 − 999 − 999 = 24. Therefore Q = 2, R = 4, and so P + Q + R = 9 + 2 + 4 = 15. Answer: (C) 18. We begin by recognizing that moving a block from Box A to Box B, and then moving the same block from Box B back to Box A has does not change the total mass in each box. Since such a pair of moves increases the number of blocks that Jasmine moves from Box A to Box B, then all such pairs of moves will not occur in counting the fewest number of blocks that Jasmine could have moved from Box A to Box B. (Similarly, moving a block from Box B to Box A, and then moving that same block from Box A back to Box B is a pair of moves that will not occur.) That is, once a block has been moved from one box to another, that block will not be moved in the opposite direction between the two boxes. Next, we consider what total masses of blocks can be moved from Box B to Box A. Box B contains one 50 g block and three 10 g blocks, and so Jasmine can move the following total masses of blocks from Box B to Box A: 10 g, 20 g, 30 g, 50 g, 60 g, 70 g, and 80 g. (Can you see how to get each of these masses using the blocks in Box B and why no other masses are possible?) If Jasmine moves 10 g from Box B to Box A, then she must move blocks whose total mass is 65 g + 10 g = 75 g from Box A to Box B so that Box A contains 65 g less than it did originally and Box B contains 65 g more than it did originally. For each of the other masses that may be moved from Box B to Box A, we summarize the total mass of blocks that Jasmine must move from Box A to Box B. Note that in each case, the mass moved from Box A to Box B must be 65 g greater than the mass moved from Box B to Box A. Mass moved from Box B to Box A 10 g 20 g 30 g 50 g 60 g 70 g 80 g Mass moved from Box A to Box B 75 g 85 g 95 g 115 g 125 g 135 g 145 g 2022 Gauss Contest Solutions Page 6 Next, we determine if it is possible for Jasmine to choose blocks from Box A whose total mass is given in the second column in the table above. Recall that Box A originally contains one 100 g block, one 20 g block, and three 5 g blocks. Is it possible for Jasmine to choose blocks from Box A whose total mass is exactly 75 g? Since the 100 g block is too heavy, and the remaining blocks have a total mass that is less than 75 g, then it is not possible for Jasmine to choose blocks from Box A whose total mass is exactly 75 g. In the table below, we determine which of the exact total masses Jasmine is able to move from Box A to Box B. For those masses which are possible, we list the number of blocks that Jasmine must move from Box A to Box B. Total mass moved from Box B to Box A 10 g 20 g 30 g 50 g 60 g 70 g 80 g Total mass moved from Box A to Box B 75 g 85 g 95 g 115 g 125 g 135 g 145 g Is it possible to move this mass from Box A to Box B? No No No Yes; one 100 g, three 5 g Yes; one 100 g, one 20 g, one 5 g Yes; one 100 g, one 20 g, three 5 g No Number of blocks moved from Box A to Box B 4 3 5 From the table above, the fewest number of blocks that Jasmine could have moved from Box A to Box B is 3. In this case, Jasmine moves a total mass of 1(100 g) + 1(20 g) + 1(5 g) = 125 g from Box A to Box B and she moves a total mass of 1(50 g) + 1(10 g) = 60 g from Box B to Box A. This confirms that Box B contains 125 g − 60 g = 65 g more than it did originally (and hence Box A contains 65 g less), as required. Answer: (A) 19. The original ratio of red candies to blue candies is 3 : 5, and so the number of red candies was a positive integer multiple of 3, and the number of blue candies was the same positive integer multiple of 5. For example, there could have been 3 × 1 = 3 red candies and 5 × 1 = 5 blue candies, or 3 × 2 = 6 red candies and 5 × 2 = 10 blue candies, or 3 × 3 = 9 red candies and 5 × 3 = 15 blue candies, and so on. We continue these possibilities in the table below, and consider the number of red and blue candies and the resulting ratio after three blue candies are removed from the dish. Possible numbers of red and blue candies originally Numbers of red and blue candies after 3 blue are removed New ratio of the number of red candies to blue 3 red, 5 blue 3 red, 2 blue 3:2 6 red, 10 blue 6 red, 7 blue 6:7 9 red, 15 blue 9 red, 12 blue 9 : 12 =3:4 12 red, 20 blue 12 red, 17 blue 12 : 17 15 red, 25 blue 15 red, 22 blue 15 : 22 18 red, 30 blue 18 red, 27 blue 18 : 27 =2:3 If the dish originally contained 18 red candies and 30 blue (note that 18 : 30 = 3 : 5), then when three blue candies are removed, the ratio of the number of red candies to blue candies 2022 Gauss Contest Solutions Page 7 becomes 18 : 27 which is equal to 2 : 3, as required. Therefore, there were 30 − 18 = 12 more blue candies than red candies in the dish before any candies were removed. Answer: (B) 20. Let Anyu, Brad, Chi, and Diego be represented by A, B, C, and D, respectively, and so their original order is ABCD. When rearranged, A is not in the 1st position, and so there are exactly 3 cases to consider: A is in the 2nd position, or A is in the 3rd position, or A is in the 4th position. For each of these 3 cases, we count the number of ways to arrange B, C, and D. Case 1: A is in the 2nd position Since A is in the 2nd position, B can be in any of the other 3 positions (1st , 3rd or 4th ). If B is in the 1st position, then there is exactly one possible rearrangement: BADC (since C and D cannot be in the 3rd and 4th positions respectively). If B is in the 3rd position, then there is exactly one possible rearrangement: DABC (since D cannot be in the 4th position). If B is in the 4th position, then there is exactly one possible rearrangement: CADB (since C cannot be in the 3rd position). Thus there are exactly 3 possible rearrangements when A is in the 2nd position. Case 2: A is in the 3rd position Since A is in the 3rd position, C can be in any of the other 3 positions. In a manner similar to Case 1, it can be shown that there are 3 possible rearrangements in this case: CDAB, DCAB, and BDAC. Case 3: A is in the 4th position Since A is in the 4th position, D can be in any of the other 3 positions. Similarly, there are 3 possible rearrangements in this case: DCBA, CDBA, and BCDA. So that each person is not in their original position, the four friends can rearrange themselves in 3 + 3 + 3 = 9 different ways. (Such a rearrangement of a list in which no element appears in its original position is called a derangement.) Answer: (B) 21. We begin by constructing the three diagonals inside the smaller squares that were missing from the diagram given in the question, as shown. The two diagonals inside each of these 4 smaller squares divide each smaller square into 4 identical triangles having equal area. Thus, square ABCD is divided into 4 × 4 = 16 such triangles. Since 7 of these triangles are shaded, then the fraction of ABCD 7 that is shaded is 16. A B D C Answer: (C) 22. Solution 1 Since the sum of p, q, r, s and the sum of q, r, s, t are both equal to 35, and q, r, s are added in each sum, then p = t. Similarly, since the sum of q, r, s, t and the sum of r, s, t, u are both equal to 35, and r, s, t are added in each sum, then q = u. 2022 Gauss Contest Solutions Page 8 It can similarly be shown that r = v and s = w. Using the above observations, the sequence p, q, r, s, t, u, v, w can be written as p, q, r, s, p, q, r, s. Since the sum of q and v is 14 and r = v, then the sum of q and r is 14. Since the sum of p, q, r, s is 35 and the sum of q and r is 14, then the sum of p and s is 35 − 14 = 21. The value of p is as large as possible exactly when the value of s is as small as possible. Since s is a positive integer, its smallest possible value is 1. Therefore, the largest possible value of p is 21 − 1 = 20. (We note that 20, 4, 10, 1, 20, 4, 10, 1 is an example of such a list.) Solution 2 The sum of the values of each group of four consecutive letters is 35. Thus, p+q+r+s = 35 and t+u+v+w = 35, and so (p+q+r+s)+(t+u+v+w) = 35+35 = 70. Rearranging the sum of these eight letters, we get p + q + r + s + t + u + v + w = (p + w) + (q + v) + (r + s + t + u) = 70 However, r + s + t + u = 35 (the sum of the values of four consecutive letters), and q + v = 14. Substituting, we get (p + w) + 14 + 35 = 70, and so p + w = 21. The value of p is as large as possible exactly when the value of w is as small as possible. Since w is a positive integer, its smallest possible value is 1. Substituting, we get p + 1 = 21, and so the largest possible value of p is 20. (We note that 20, 12, 2, 1, 20, 12, 2, 1 is an example of such a list.) Answer: (C) 23. Katharina placed the 8 letters in a mixed-up order around the circle. 1 We number the position of L at the top the circle as 1, the next 8 2 number moving clockwise from L position 2, and so on through to position 8, as shown. 3 7 Jaxon begins at position 1, writes down the letter L, and moves clockwise writing down every third letter that he has not yet 6 4 written down. 5 Thus, the first 3 letters that Jaxon writes down are the letters in positions 1, 4 and 7. Continuing to move clockwise around the circle, the next three positions at which letters have not yet been written down are positions 8, 2 and 3 (the letter in position 1 was already written down), and so Jaxon writes down the letter in position 3. The next three positions at which letters have not yet been written down are 5, 6 and 8 (the letters in positions 4 and 7 were already written down), and so Jaxon writes down the letter in position 8. At this point, Jaxon has written the letters in positions 1, 4, 7, 3, and 8. The next three positions at which letters have not yet been written down are 2, 5 and 6 (the letters in positions 1, 3 and 4 were already written down), and so Jaxon writes down the letter in position 6. At this point, only the letters in positions 2 and 5 have not been written. Since Jaxon left off at position 6, he skips the letter at position 2, skips the letter at position 5, and then writes down the letter at position 2. Finally, Jaxon writes the final letter at position 5. Thus in order, Jaxon writes the letters in positions 1, 4, 7, 3, 8, 6, 2, and 5. 2022 Gauss Contest Solutions Page 9 Since Jaxon’s list is L, M, N, O, P, Q, R, S, then L is the letter in position 1 of Katharina’s ordering, M is the letter in position 4, N is the letter in position 7, and so on. Therefore, Katharina’s clockwise order is L, R, O, M, S, Q, N, P. Answer: (C) 24. A palindrome greater than 10 000 and less than 100 000 is a 5-digit positive integer of the form abcba, where a, b and c are digits and a 6= 0. A positive integer is a multiple of 18 if it is a multiple of both 2 and 9 (and a positive integer that is a multiple of both 2 and 9 is a multiple of 18). A positive integer is a multiple of 2 if it is even, and thus the digit a is equal to 2, 4, 6 or 8 (recall a 6= 0). A positive integer is a multiple of 9 exactly when the sum of its digits is a multiple of 9, and thus a + b + c + b + a or 2a + 2b + c is a multiple of 9. Next we consider four possible cases, one case for each of the possible values of a. Case 1: a = 2 When a = 2, we require that 2a + 2b + c = 4 + 2b + c be a multiple of 9. Since 4 + 2b + c ≥ 4, then the smallest possible multiple of 9 that 4 + 2b + c can equal is 9. Since b ≤ 9 and c ≤ 9, then 4 + 2b + c is at most 4 + 2(9) + 9 = 31. Thus, 4 + 2b + c can equal 9, 18 or 27, which gives 2b + c equal to 5, 14 or 23 respectively. Next, we determine the possible values of b and c so that 2b + c is equal to 5, 14 or 23. 2b + c = 5 b = 2, c = 1 b = 1, c = 3 b = 0, c = 5 2b + c = 14 b = 7, c = 0 b = 6, c = 2 b = 5, c = 4 b = 4, c = 6 b = 3, c = 8 2b + c = 23 b = 9, c = 5 b = 8, c = 7 b = 7, c = 9 Thus when a = 2, there are 3 + 5 + 3 = 11 such palindromes. Case 2: a = 4 When a = 4, we require that 2a + 2b + c = 8 + 2b + c be a multiple of 9. Since 8 + 2b + c ≥ 8, then the smallest possible multiple of 9 that 8 + 2b + c can equal is 9. Since b ≤ 9 and c ≤ 9, then 8 + 2b + c is at most 8 + 2(9) + 9 = 35. Thus, 8 + 2b + c can equal 9, 18 or 27, which gives 2b + c equal to 1, 10 or 19 respectively. Next, we determine the possible values of b and c so that 2b + c is equal to 1, 10 or 19. 2b + c = 1 b = 0, c = 1 2b + c = 10 b = 5, c = 0 b = 4, c = 2 b = 3, c = 4 b = 2, c = 6 b = 1, c = 8 2b + c = 19 b = 9, c = 1 b = 8, c = 3 b = 7, c = 5 b = 6, c = 7 b = 5, c = 9 Thus when a = 4, there are 1 + 5 + 5 = 11 such palindromes. Case 3: a = 6 When a = 6, we require that 2a + 2b + c = 12 + 2b + c be a multiple of 9. Since 12 + 2b + c ≥ 12 and 12 + 2b + c ≤ 12 + 2(9) + 9 = 39, then 12 + 2b + c can equal 18, 27 or 36, which gives 2b + c equal to 6, 15 or 24, respectively. 2022 Gauss Contest Solutions Page 10 2b + c = 6 b = 3, c = 0 b = 2, c = 2 b = 1, c = 4 b = 0, c = 6 2b + c = 15 b = 7, c = 1 b = 6, c = 3 b = 5, c = 5 b = 4, c = 7 b = 3, c = 9 2b + c = 24 b = 9, c = 6 b = 8, c = 8 Thus when a = 6, there are 4 + 5 + 2 = 11 such palindromes. Case 4: a = 8 When a = 8, we require that 2a + 2b + c = 16 + 2b + c be a multiple of 9. Since 16 + 2b + c ≥ 16 and 16 + 2b + c ≤ 16 + 2(9) + 9 = 43, then 16 + 2b + c can equal 18, 27 or 36, which gives 2b + c equal to 2, 11 or 20, respectively. 2b + c = 2 b = 1, c = 0 b = 0, c = 2 2b + c = 11 b = 5, c = 1 b = 4, c = 3 b = 3, c = 5 b = 2, c = 7 b = 1, c = 9 2b + c = 20 b = 9, c = 2 b = 8, c = 4 b = 7, c = 6 b = 6, c = 8 Thus when a = 8, there are 2 + 5 + 4 = 11 such palindromes. Therefore, the number of palindromes that are greater than 10 000 and less than 100 000 and that are multiples of 18 is 11 + 11 + 11 + 11 = 44. Answer: (D) 25. After all exchanges there are 4 balls in each bag, and so if a bag contains exactly 3 different colours of balls, then it must contain exactly 2 balls of the same colour and 2 balls each having a colour that is different than all other balls in the bag. Among the 8 balls in the two bags, there are 2 red balls and 2 black balls and each of the remaining balls has a colour that is different than all the other balls. Thus after all exchanges, one bag must contain the 2 red balls and the other bag must contain the 2 black balls. Since Becca’s bag initially contains both black balls, and Becca moves only 1 ball from her bag to Arjun’s bag, it is not possible for Arjun’s bag to contain the 2 black balls after all exchanges. This tells us that if each bag contains exactly 3 different colours of balls after all exchanges, then Arjun’s bag contains the 2 red balls and Becca’s bag contains the 2 black balls. We let the first letter of each colour represent a ball of that colour. Then initially, Arjun’s bag contains RRGY V and Becca’s bag contains BBO. For the first ball chosen to be moved, there are exactly two cases to consider: Case 1: The first ball moved from Arjun’s bag to Becca’s bag is R, or Case 2: The first ball moved from Arjun’s bag to Becca’s bag is not R (thus it is G, Y or V ). We begin with Case 1 and determine the probability that after all exchanges, each bag contains exactly 3 different colours of balls. Case 1: The first ball moved from Arjun’s bag to Becca’s bag is R Since Arjun’s bag initially contains 5 balls, 2 of which are R, the probability that R is chosen as the first ball to move is 25. After R is moved from Arjun’s bag to Becca’s, Arjun’s bag contains RGY V and Becca’s bag 2022 Gauss Contest Solutions Page 11 contains BBOR. Since Arjun’s bag must contain both R’s after all exchanges, and Becca moves only 1 ball from her bag to Arjun’s, then the next ball chosen to move must be R. Becca’s bag contains 4 balls, 1 of which is R, and so in this case the probability that R is chosen as the second ball to move is 14. After the first two balls are moved, Arjun’s bag contains RRGY V and Becca’s bag contains BBO. Finally, a ball is chosen from Arjun’s bag and moved to Becca’s. Since Arjun’s bag must contain both R’s after all exchanges, then the ball that is chosen must be G, Y or V which occurs with probability 53. If the first ball moved from Arjun’s bag to Becca’s bag is R, then the probability that each bag contains exactly 3 different colours of balls after all exchanges is given by the product of the 6 3 individual probabilities of choosing each of the 3 balls or 52 × 41 × 35 = 100 = 50. Case 2: The first ball moved from Arjun’s bag to Becca’s bag is G, Y or V Since Arjun’s bag initially contains 5 balls, the probability that G, Y or V is chosen as the first ball to move is 53. After one of G, Y or V is moved from Arjun’s bag to Becca’s, Arjun’s bag contains RR and two of G, Y and V , and Becca’s bag contains BBO and one of G, Y or V. For the second ball chosen to be moved, there are exactly two cases to consider and so we split Case 2 into these two separate cases. Case 2a: The first ball moved from Arjun’s bag to Becca’s bag is G, Y or V , and the ball moved from Becca’s bag to Arjun’s bag is B, or Case 2b: The first ball moved from Arjun’s bag to Becca’s bag is G, Y or V , and the ball moved from Becca’s bag to Arjun’s bag is not B (that is, it is O or the first ball moved). We begin with Case 2a and determine the probability that after all exchanges, each bag contains exactly 3 different colours of balls. Case 2a: The first ball moved from Arjun’s bag to Becca’s bag is G, Y or V , and the ball moved from Becca’s bag to Arjun’s bag is B We previously determined that the probability that G, Y or V is chosen as the first ball to move is 35. At this point, Becca’s bag contains 4 balls, 2 of which are B’s, and so the probability that B is chosen is 42 = 12. After B is moved from Becca’s bag to Arjun’s, Arjun’s bag contains RRB and two of G, Y and V and Becca’s bag contains BO and one of G, Y or V. Since Becca’s bag must contain both B’s after all exchanges, then the final ball chosen to move must be B. Arjun’s bag contains 5 balls, 1 of which is B, and so in this case the probability that B is chosen as the final ball to move is 51. If the first ball moved from Arjun’s bag to Becca’s bag is G, Y or V , and the ball moved from Becca’s bag to Arjun’s bag is B, then the probability that each bag contains exactly 3 different colours of balls after all exchanges is given by the product of the individual probabilities of 3 choosing each of the 3 balls or 53 × 21 × 15 = 50. Case 2b: The first ball moved from Arjun’s bag to Becca’s bag is G, Y or V , and the ball moved from Becca’s bag to Arjun’s bag is O or the first ball moved (G, Y or V ) The probability that G, Y or V is chosen as the first ball to move is 35. At this point, Becca’s bag contains 4 balls, 2 of which are B, and so the probability that B is 2022 Gauss Contest Solutions Page 12 not chosen is 24 = 12. After this ball (O, G, Y or V ) is moved from Becca’s bag to Arjun’s, Arjun’s bag contains RR and three of O, G, Y and V and Becca’s bag contains two B’s and one of O, G, Y or V. Since Arjun’s bag must contain both R’s after all exchanges, then the final ball chosen to move must be one of the 3 balls in Arjun’s bag that is not R. Arjun’s bag contains 5 balls, 2 of which are R, and so in this case the probability that R is not chosen as the final ball to move is 35. If the first ball moved from Arjun’s bag to Becca’s bag is G, Y or V , and the ball moved from Becca’s bag to Arjun’s bag is not B, then the probability that each bag contains exactly 3 different colours of balls after all exchanges is given by the product of the individual probabilities 9 of choosing each of the 3 balls or 53 × 21 × 35 = 50. Finally, the probability that each bag contains exactly 3 different colours of balls after all 3 3 9 3 exchanges is 50 + 50 + 50 = 15 = 10. 50 Answer: (A) 2022 Gauss Contest Solutions Page 13 Grade 8 1. The regular pentagon shown has 5 sides, each with length 2 cm. The perimeter of the pentagon is 5 × 2 cm = 10 cm. Answer: (E) 2. The faces shown are labelled with 1, 3 and 5 dots. Therefore, the other three faces are labelled with 2, 4 and 6 dots. The total number of dots on the other three faces is 2 + 4 + 6 = 12. Answer: (D) 3. If the number is n, then “a number increased by five” is best represented by the expression n + 5. The equation that best represents “a number increased by five equals 15” is n + 5 = 15. Answer: (C) 4. Reading from the graph, the approximate number of bobbleheads sold in the years 2016, 2017, 2018, 2019, 2020, and 2021 were 20, 35, 40, 38, 60, and 75, respectively. Beginning with 2016 and 2017, the increases (or decreases) in the sale of bobbleheads between consecutive years are approximately 35 − 20 = 15, 40 − 35 = 5, 38 − 40 = −2 (a decrease of 2), 60 − 38 = 22, and 75 − 60 = 15. Thus the greatest increase in the sale of bobbleheads was approximately 22 and this occurred between 2019 and 2020. Answer: (D) 5. Continuing to count down by 11, we get 72, 61, 50, 39, 28, 17, 6, −5,... The last number that Aryana will count that is greater than 0 is 6. Answer: (C) 6. Since ∠ABC = 90◦ , then 44 + x + x = 90 or 2x = 90 − 44, and so 2x = 46 or x = 23. Answer: (B) 7. Since 45 = 1 41 and 12 = 1, then each of the choices (A) −1, (B) 54 and (C) 12 is at least 1 away from 0. Since − 45 = −0.8 and −0.8 is closer to 0 than 0.9 is to 0, then − 54 is the closest value to zero. Answer: (D) 8. Solution 1 The value of 4 quarters is $1.00. Thus, the value of 4 × 100 = 400 quarters is $1.00 × 100 = $100.00. Since the jar initially contains 267 quarters, then 400 − 267 = 133 quarters must be added to the jar for the total value of the quarters to equal $100.00. Solution 2 The value of 267 quarters is $0.25 × 267 = $66.75 and so $100.00 − $66.75 = $33.25 must be added to the jar so that it contains $100.00. $33.25 Thus, the number of quarters that must be added to the jar is = 133. $0.25 Answer: (D) 2022 Gauss Contest Solutions Page 14 9. Each package of greeting cards comes with 10 − 8 = 2 more envelopes than cards. Thus, 3 packages of greeting cards comes with 3 × 2 = 6 more envelopes than cards, and 4 packages of greeting cards comes with 4 × 2 = 8 more envelopes than cards. Kirra began with 7 cards and no envelopes. To have more envelopes than cards, Kirra must buy enough packages to make up the difference between the number of cards and the number of envelopes (which is 7). Thus, the smallest number of packages that Kirra must buy is 4. Note: We can check that if Kirra buys 3 packages she has 3 × 8 + 7 = 31 cards and 3 × 10 = 30 envelopes, and thus fewer envelopes than cards. However, if she buys 4 packages, she has 4 × 8 + 7 = 39 cards and 4 × 10 = 40 envelopes, and thus more envelopes than cards, as required. Answer: (B) 10. The horizontal distance between the point (a, b) and the y-axis is a units, and the horizontal distance between the point (c, d) and the y-axis is c units. Since (a, b) is horizontally farther from the y-axis than (c, d) is from the y-axis, then a > c and so statement (E) is true. Let us consider why each of the other statements is false. The points (a, b) and (c, d) each lie above the x-axis and thus b > 0 and d > 0. The point (e, f ) lies below the x-axis and thus f < 0. Since b > 0 and f < 0, then b > f and so statement (C) is false. Further, the vertical distance between the point (a, b) and the x-axis is b units, and the vertical distance between the point (c, d) and the x-axis is d units. Since (a, b) is vertically farther from the x-axis than (c, d) is from the x-axis, then b > d and so statement (B) is false. Similarly, the points (a, b) and (c, d) each lie to the right of the y-axis and thus a > 0 and c > 0. The point (e, f ) lies to the left of the y-axis and thus e < 0. Since a > 0 and e < 0, then a > e and so statement (D) is false. Since c > 0 and e < 0, then c > e and so statement (A) is also false. Answer: (E) 11. In the sequence, the letters of the alphabet repeat in blocks of 26 letters. Thus, 10 of these blocks gives a sequence that contains 10 × 26 = 260 letters. Each block of 26 letters ends with the letter Z, and so the 260th letter is a Z. Moving backward in the sequence from this Z, we get that the 259th letter is a Y , and the 258th letter is an X. Answer: (C) 12. The first Wednesday of a month must occur on one of the first 7 days of a month, that is, the 1st , 2nd , 3rd , 4th , 5th , 6th , or 7th of the month. The second Wednesday of a month occurs 7 days following the first Wednesday of that month, and the third Wednesday of a month occurs 14 days following the first Wednesday of that month. Adding 14 days to each of the possible dates for the first Wednesday of the month, we get that the third Wednesday of a month must occur on the 15th , 16th , 17th , 18th , 19th , 20th , or 21st of that month. Of the given answers, the public holiday cannot occur on the 22nd of that month. Answer: (B) 2022 Gauss Contest Solutions Page 15 13. The probability that the arrow stops on the largest section is 50% or 12. The probability that it stops on the next largest section is 1 in 3 or 31. Thus, the probability that the arrow stops on the smallest section is 1 − 12 − 13 = 66 − 36 − 62 = 16. Answer: (C) 14. A positive number is divisible by both 3 and 4 if it is divisible by 12 (and a positive number that is divisible by 12 is divisible by both 3 and 4). The positive two-digit numbers that are divisible by 12 (and thus 3 and 4) are 12, 24, 36, 48, 60, 72, 84, and 96. Of these, 60, 72, 84, and 96 have a tens digit that is greater than its ones digit. Thus, there are 4 positive two-digit numbers that satisfy the given property. Answer: (A) 15. Solution 1 The area of the walkway is equal to the area of the pool subtracted from the combined area of the pool and walkway. That is, if the area of the walkway is Aw and the area of the pool is Ap , then Aw = (Aw +Ap )−Ap. The combined area of the pool and walkway is equal to the area of the rectangle with length 22 m and width 10 m. The length, 22 m, is given by the 20 m pool length plus the 1 m wide walkway on each end of the pool. Similarly, the 10 m width is given by the 8 m pool width plus the 1 m wide walkway on each side of the pool. Thus the area of the walkway is Aw = (Aw + Ap ) − Ap = (22 m × 10 m) − (20 m × 8 m) = 220 m2 − 160 m2 = 60 m2 Solution 2 We begin by extending each side of the pool 1 m in each direction. This divides the area of the walkway into four 1 m by 1 m squares (in the corners), and two 20 m by 1 m rectangles (along the 20 m sides of the pool), and two 8 m by 1 m rectangles (along the 8 m sides of the pool), as shown. Thus the area of the walkway is 8m 20 m 1m 1m 4 × (1 m × 1 m) + 2 × (20 m × 1 m) + 2 × (8 m × 1 m) = 4 m2 + 40 m2 + 16 m2 = 60 m2 Answer: (B) 16. Reading from the Venn diagram, 5 students participate in both music and sports, 15 students participate in music (and not sports), and 20 students participate in sports (and not music). Thus, there are 5 + 15 + 20 = 40 students that participate in music or sports or both, and so there are 50 − 40 = 10 students that do not participate in music and do not participate in sports. 10 Of the 50 students, 50 × 100% = 20% do not participate in music and do not participate in sports. Answer: (C) 2022 Gauss Contest Solutions Page 16 17. If the number of golf balls in Bin G is x, then the number of golf balls in Bin F is 23 x. In this case, the total number of golf balls is x + 32 x = 33 x + 23 x = 53 x, and so 35 x = 150. Multiplying both sides of this equation by 3, we get 5x = 150 × 3 = 450. Dividing both sides by 5, we get x = 450 = 90 and so there are 90 golf balls in Bin G. 5 The number of golf balls in Bin F is 150 − 90 = 60, and so there are 90 − 60 = 30 fewer golf balls in Bin F than in Bin G. (Alternately, we may have begun by letting there be 3x golf balls in Bin G and 2x golf balls in Bin F.) Answer: (B) 18. Figure 1 is formed with 7 squares. Figure 2 is formed with 5 + 7 squares. Figure 3 is formed with 5 + 5 + 7 = 2 × 5 + 7 squares. Figure 4 will be formed with 5 + 5 + 5 + 7 = 3 × 5 + 7 squares. Figure 5 will be formed with 5 + 5 + 5 + 5 + 7 = 4 × 5 + 7 squares. Thus, the number of groups of 5 squares needed to help form each figure is increasing by 1. Also, in each case the number of groups of 5 squares needed is one less than the Figure number. For example, Figure 6 will be formed with 5 groups of 5 squares plus 7 additional squares. In general, we can say that Figure N will be formed with N − 1 groups of 5 squares, plus 7 additional squares. Since 2022 = 403 × 5 + 7, the figure with 2022 squares has 403 groups of 5 squares, plus 7 additional squares. Thus, N − 1 = 403 and so N = 404. The number of the figure that has 2022 squares is 404. Answer: (C) 19. There are 60 minutes in an hour, and so the number of minutes between 7 a.m. and 11 a.m. is 4 × 60 = 240. Since Mateo stopped for a 40 minute break, he drove for 240 − 40 = 200 minutes. 300 km = 1.5 km/min. Thus, the average speed for Mateo’s 300 km trip was 200 minutes Since there are 60 minutes in an hour, if Mateo averaged 1.5 km per minute, then he averaged 1.5 × 60 km per hour or 90 km/h. Answer: (C) A 20. Since 4ABC is equilateral and has sides of length 4, then AB = BC = AC = 4. 4 The midpoint of BC is D and so BD = CD = 2. E The midpoint of AD is E and so AE = ED. Since AB = AC and D is the midpoint of BC, then AD is B C 2 D 2 perpendicular to BC, as shown. Triangle ADC is a right-angled triangle, and so by the Pythagorean Theorem, we get (AC)2 = (AD)2 + (DC)2 or 42 = (AD)2 + 22 , and so (AD)2 = 16 − 4 = 12. Similarly, 4EDC is right-angled, and so by the Pythagorean Theorem, we get (EC)2 = (ED)2 + (DC)2 or (EC)2 = (ED)2 + 22. Since ED = 21 AD, then (ED)2 = 12 AD × 12 AD or (ED)2 = 14 (AD)2. Since AD2 = 12, then (ED)2 = 41 × 12 = 3. Substituting, we get (EC)2 = 3 + 22 , and so (EC)2 = 7. Answer: (A) 2022 Gauss Contest Solutions Page 17 21. A perfect square is a number that can be expressed as the product of two equal integers. By this definition, 0 is a perfect square since 0 × 0 = 0. Since the product of 0 and every positive integer is 0, then every positive integer is a factor of 0, and so 0 has an infinite number of positive factors. The next three smallest perfect squares are 12 = 1, 22 = 4 and 32 = 9. Each of these has at most three positive factors. The next largest perfect square is 42 = 16. The positive factors of 16 are 1, 2, 4, 8, and 16, and so 16 is a perfect square that has exactly five positive factors. The remaining perfect squares that are less than 100 are 25, 36, 49, 64, and 81. Both 25 and 49 each have exactly three positive factors. The positive factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. The positive factors of 64 are 1, 2, 4, 8, 16, 32, and 64. Thus, both 36 and 64 each have more than five positive factors. Finally, the positive factors of 81 are 1, 3, 9, 27, and 81. The two perfect squares that are less than 100 and that have exactly five positive factors are 16 and 81, and their sum is 16 + 81 = 97. Answer: (E) 22. The sum of the values of each group of three consecutive letters is 35. Thus, r + s + t = 35 and s + t + u = 35. Since each of these equations is equal to 35, then the left sides of the two equations are equal to each other. That is, r + s + t = s + t + u and since s + t is common to both sides of this equation, then r = u. Since q + u = 15 and r = u, then q + r = 15. Since p + q + r = 35 and q + r = 15, then p = 35 − 15 = 20. Finally, we get p + q + r + s + t + u + v = p + (q + r + s) + (t + u + v) = 20 + 35 + 35 = 90 Answer: (D) 23. Consider folding the net into the cube and positioning the cube E A with the face labelled F down (on the bottom), and so the face D labelled A is on top and the four remaining vertical faces are as B C shown. Beginning at A, the ant has 4 choices for which face to visit next. F That is, the ant can walk from A to any of the vertical faces, B, C, D, or E. From each of these vertical faces, the ant can walk to F (the bottom face), or the ant can walk to an adjacent vertical face. We call these two possibilities Case 1 and Case 2, and for each case we consider the number of possible orders in which the ant can visit the faces. Case 1: The ant’s 2nd move is to F , the bottom face We begin by recognizing that there is only 1 choice for the ant’s 2nd move in this case. From F , the ant’s 3rd move must be back to a vertical face. The ant cannot return to the vertical face that it has already visited. Also, the ant cannot move to the vertical face that is opposite the vertical face it has already 2022 Gauss Contest Solutions Page 18 visited. Why? Consider for example that the ant visits, in order, A, B, F. If the ant walks to E (the face opposite B) on its 3rd move, then its 4th move must be to C or to D (since it has visited the other four faces). However, once at C or D, the ant is “trapped” since it has already visited all four adjacent faces, and thus cannot get to the sixth face. Thus from F , the ant’s 3rd move must be to a vertical face that is not the face already visited, and is not the face opposite the face already visited, and so there are 2 choices for the next move. From this vertical face, the ant must walk to another vertical face (since it has already visited A and F ). One such adjacent face has already been visited and the other has not, and so the ant’s 1 choice is for its 4th move to be to the adjacent vertical face it has not visited. For example, if the order is A, B, F, C, then the ant must move to E next. Finally, the ant’s final move must be to the adjacent vertical face that it has not visited. Summarizing Case 1, there are 4 choices for the first move from A to one of the vertical faces, 1 choice for the 2nd move (to F ), 2 choices for the 3rd move, and 1 choice for each of the last two moves, and so there are 4 × 1 × 2 × 1 × 1 = 8 possible orders. Case 2: The ant’s 2nd move is to an adjacent vertical face There are two vertical faces adjacent to each vertical face, and so the ant has 2 choices for its 2nd move. For its 3rd move, the ant can walk to the bottom face F , or the ant can walk to the adjacent face that has not been visited. We call the first of these Case 2a and the second Case 2b. Case 2a: The ant’s 3rd move is to F We begin by recognizing that there is only 1 choice for the ant’s 3rd move in this case. From F , the ant’s 4th move must be back to a vertical face. The ant cannot return to either of the two vertical faces that it has already visited, and so there are 2 choices for the ant’s 4th move. For example, if the order is A, B, C, F , the ant’s 4th move can be to D or to E. The ant’s final move is to the final vertical face and thus there is only 1 choice. Summarizing Case 2a, there are 4 choices for the first move from A to one of the vertical faces, 2 choices for the 2nd move (to an adjacent vertical face), 1 choice for the 3rd move (to F ), 2 choices for the 4th move, and 1 choice for the last move, and so there are 4 × 2 × 1 × 2 × 1 = 16 possible orders. Case 2b: The ant’s 3rd move is to the adjacent face that has not been visited We begin by recognizing that there is only 1 choice for the ant’s 3rd move in this case. At this point the ant has visited three vertical faces. The ant’s 4th move can be to F or to the final vertical face. That is, the ant has 2 choices for its 4th move. If the ant’s 4th move is to F , then its last move is to the remaining vertical face. If the ant’s 4th move is to the final vertical face, then its last move is to F. That is, once the ant chooses its 4th move, it has only 1 choice for its final move. Summarizing Case 2b, there are 4 choices for the first move from A to one of the 2022 Gauss Contest Solutions Page 19 vertical faces, 2 choices for the 2nd move (to an adjacent vertical face), 1 choice for the 3rd move (to the adjacent vertical face), 2 choices for the 4th move (to F or to the final vertical face), and 1 choice for the last move, and so there are 4×2×1×2×1 = 16 possible orders. Thus, if the ant starts at A and visits each face exactly once, there are 8 + 16 + 16 = 40 possible orders. Answer: (E) 24. Solution 1 We begin by recognizing that numbers with the given property cannot have two digits that are zero. Can you see why? Thus, numbers with this property have exactly one zero or they have no zeros. We consider each of these two cases separately. Case 1: Suppose the number has exactly one digit that is a zero. Each of the numbers greater than 100 and less than 999 is a three-digit number and so in this case, the number also has two non-zero digits. Since one of the digits is equal to the sum of the other two digits and one of the digits is zero, then the two non-zero digits must be equal to one another. The two non-zero digits can equal any integer from 1 to 9, and thus there are 9 possible values for the non-zero digits. For each of these 9 possibilities, the zero digit can be the second digit in the number or it can be the third digit (the first digit cannot be zero). That is, there are 9 possible values for the non-zero digits and 2 ways to arrange the three digits, and thus 9 × 2 = 18 numbers of this form satisfy the given property. For example, numbers of this form are 101 and 110, 202 and 220, and so on. Case 2: Suppose the number has no digits that are equal to zero. Let the three digits of the number be a, b and c, arranged in some order. Assume that a is the largest digit, and so a = b + c. If a = b, then c = 0 which contradicts our assumption that no digit is equal to 0. Similarly, if a = c, then b = 0 and the same contradiction arises. Thus a > b and a > c. If a = 1, then b = c = 0 (since a > b and a > c), but then a 6= b + c. Therefore, a is at least 2. If a = 2, then b = c = 1 and these are the only possible values for b and c when a = 2. In this case, the 3 ways to arrange these digits give the numbers 112, 121 and 211, each with the desired property. If a = 3, then b and c equal 1 and 2, in some order. In this case, the 6 ways to arrange these digits give the numbers 123, 132, 213, 231, 312, and 321, each with the desired property. From these cases, we recognize that if b = c, then there are 3 possible arrangements of the digits. However, if the three digits are different from one another, then there are 3 choices for the first digit, 2 choices for second digit and 1 choice for the third digit, and thus 3 × 2 × 1 = 6 ways to arrange the digits. We consider all possible values of a, b, c and count the arrangements of these digits in the table below. 2022 Gauss Contest Solutions Values for a 2 3 4 5 6 7 8 9 Page 20 Values for b and c with the number of arrangements in brackets [ ] 1,1 1, 2 1, 3 2, 2 1, 4 2, 3 1, 5 2, 4 3, 3 1, 6 2, 5 3, 4 1, 7 2, 6 3, 5 4, 4 1, 8 2, 7 3, 6 4, 5 Total number of arrangements 3 6 9 12 15 18 21 24 Thus, there are 18 + 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 = 126 numbers that satisfy the given property. Solution 2 We begin by recognizing that numbers with the given property cannot have three equal digits. Can you see why? Thus, numbers with this property have exactly two digits that are equal or all three digits are different. We consider each of these two cases separately. Case 1: Suppose the number has exactly two digits that are equal. The two equal digits cannot be 0 since then the third digit would also be 0. If for example the two equal digits are 1s, there are two possibilities for the third digit. The third digit can be 2 (since 1 + 1 = 2) or the third digit can be 0 (since 1 + 0 = 1). In the cases for which the third digit is equal to the sum of the two equal digits, the equal digits can be 1, 2, 3, or 4 and the third digit is 2, 4, 6, 8, respectively. (We note that the equal digits cannot be greater than 4 since their sum is greater than 9.) For each of these 4 possibilities, there are 3 ways to arrange the digits. For example, when the equal digits are 1s and the third digit is 2, the numbers 112, 121 and 211 have the desired property. Thus, there are 4 × 3 = 12 such numbers for which the third digit is equal to the sum of the two equal digits. In the cases for which two of the digits are equal and the third digit is 0, the equal digits can be any integer from 1 to 9 inclusive. For each of these 9 possibilities, there are 2 ways to arrange the digits. For example, when the equal digits are 1s and the third digit is 0, the numbers 101 and 110 have the desired property. Thus, there are 9 × 2 = 18 numbers which have two equal digits and the third digit is 0. In total, there are 12 + 18 = 30 numbers which have two equal digits and satisfy the given property. Case 2: Suppose all three digits are different from one another. Let the three digits of the number be a, b and c, arranged in some order with a > b > c. Since a is the largest digit, then a = b + c. If a = 1, then b = c = 0 (since a > b and a > c), but this is not possible since b > c. Similarly, if a = 2, then b = 1 and c = 0, however these digits do not satisfy the given property. Therefore, a is at least 3. If a = 3, then b = 2 and c = 1. 2022 Gauss Contest Solutions Page 21 In this case, the 6 ways to arrange these digits give the numbers 123, 132, 213, 231, 312, 321 and each has the desired property. We consider all possible values of a, b, c in the table below. Values for a 3 4 5 6 7 8 9 Values for b, c 2, 1 3, 1 4, 1 5, 1 6, 1 7, 1 8, 1 3, 2 4, 2 5, 2 6, 2 7, 2 4, 3 5, 3 6, 3 5, 4 For each of these 16 possibilities in the table above, there are 6 ways to arrange the three digits, and so there are 16 × 6 = 96 such numbers. Thus, there are a total of 30 + 96 = 126 numbers that satisfy the given property. Answer: (B) 25. Of the 4200 samples to test, let the number of samples that contain blueberry be x. Since each sample either contains blueberry or it does not, then there are 4200 − x samples that do not contain blueberry. Student A reports correctly on 90% of the x samples that contain blueberry. 90 Thus, Student A reports 100 x of these samples contain blueberry. Student A reports correctly on 88% of the 4200 − x samples not containing blueberry and thus incorrectly reports that 100% − 88% = 12% of these samples contain blueberry. (That is, when a student is wrong when reporting “no blueberry”, it means that there is blueberry.) 12 Therefore, Student A reports 100 (4200 − x) of these samples contain blueberry. In total, Student A reports that 90 12 x + 100 (4200 − x) samples contain blueberry. 100 98 14 x+ 100 (4200−x) samples contain blueberry, and 100 Similarly, Student B reports that Student C 2m 100−4m reports that 100 x + (4200 − x) samples contain blueberry. 100 Student B reports 315 more samples as containing blueberry than Student A, and so 14 90 12 98 x + 100 (4200 − x) − 100 x + 100 (4200 − x) = 315 100 Clearing fractions by multiplying by 100, the equation becomes 98x + 14(4200 − x) − (90x + 12(4200 − x)) = 31 500 Solving this equation, we get 98x + 14(4200 − x) − (90x + 12(4200 − x)) 98x + 58 800 − 14x − (90x + 50 400 − 12x) 98x + 58 800 − 14x − 90x − 50 400 + 12x 6x + 8400 6x x = = = = = = 31 500 31 500 31 500 31 500 23 100 3850 2022 Gauss Contest Solutions Page 22 Thus, there are 3850 samples that contain blueberry and 4200 − 3850 = 350 samples that do not contain blueberry. Together, the three students report 98 14 90 12 2m 100−4m (3850) + (350) + (3850) + (350) + (3850) + (350) 100 100 100 100 100 100 samples as containing blueberry. Simplifying this expression, we get 98 14 90 12 (3850) + 100 (350) + 100 (3850) + 100 (350) + 100 = 3773 + 49 + 3465 + 42 + 77m + 350 − 14m = 7679 + 63m 2m (3850) 100 + 100−4m 100 (350) For some positive integers m, together the three students report 7679 + 63m samples as containing blueberry. If 7679 + 63m is greater than 8000, then 63m is greater than 8000 − 7679 = 321 and so m is greater than 321 ≈ 5.09. 63 Since m is a positive integer, then m is greater than or equal to 6. Similarly, if 7679 + 63m is less than 9000, then 63m is less than 9000 − 7679 = 1321 and so m ≈ 20.97. is less than 1321 63 Since m is a positive integer, then m is less than or equal to 20. Thus, we want all integers m from 6 to 20 inclusive for which 7679 + 63m is equal to a multiple of 5. An integer is a multiple of 5 exactly when its units (ones) digit is 0 or 5. The units digit of the total number of samples, 7679 + 63m, is determined by the sum of the units digits of 7679, which is 9, and the units digit of the value of 63m. Thus, 7679 + 63m is a multiple of 5 exactly when the value of 63m has a units digit of 1 or 6 (since 9 + 1 has a units digit of 0 and 9 + 6 has a units digit of 5). The value of 63m has a units digit of 1 exactly when m has a units digit of 7 (since 3 × 7 has a units digit of 1). The value of 63m has a units digit of 6 exactly when m has a units digit of 2 (since 3 × 2 has a units digit of 6). The values of m from 6 to 20 inclusive having a units digit of 7 or 2 are 7, 12 and 17. We can confirm that when m is equal to 7, 12 and 17, the values of 7679 + 63m are 8120, 8435 and 8750 respectively, as required. The sum of all values of m for which the total number of samples that the three students report as containing blueberry is a multiple of 5 between 8000 and 9000 is 7 + 12 + 17 = 36. Answer: (B) The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2021 Gauss Contests (Grades 7 and 8) Wednesday, May 12, 2021 (in North America and South America) Thursday, May 13, 2021 (outside of North America and South America) Solutions ©2021 University of Waterloo 2021 Gauss Contest Solutions Page 2 Centre for Education in Mathematics and Computing Faculty and Staff Ed Anderson Jeff Anderson Terry Bae Jacquelene Bailey Shane Bauman Jenn Brewster Ersal Cahit Diana Castañeda Santos Sarah Chan Ashely Congi Serge D’Alessio Fiona Dunbar Mike Eden Sandy Emms Barry Ferguson Steve Furino Lucie Galinon Robert Garbary Rob Gleeson Sandy Graham Conrad Hewitt Angie Hildebrand Carrie Knoll Wesley Korir Judith Koeller Laura Kreuzer Bev Marshman Paul McGrath Jen Nelson Ian Payne J.P. Pretti Alexandra Rideout Nick Rollick Kim Schnarr Ashley Sorensen Ian VanderBurgh Troy Vasiga Christine Vender Heather Vo Bonnie Yi Gauss Contest Committee Ashley Sorensen (Chair), University of Waterloo, Waterloo, ON Kevin Grady (Assoc. Chair), Cobden District P.S., Cobden, ON Sarah Garrett, Mitchell Woods P.S., Guelph, ON Kora Lee Gallant, Madeline Symonds M.S., Hammonds Plains, NS JoAnne Halpern, Toronto, ON Clay Kellough, Fort Richmond C.I., Winnipeg, MB David Matthews, Waterloo, ON John McLoughlin, University of New Brunswick, Fredericton, NB Nick Rollick, University of Waterloo, Waterloo, ON David Switzer, Scott Central P.S., Sandford, ON Rachael Verbruggen, University of Waterloo, Waterloo, ON Robert Wong, Edmonton, AB Chris Wu, Ledbury Park E. and M.S., Toronto, ON Lori Yee, William Dunbar P.S., Pickering, ON 2021 Gauss Contest Solutions Page 3 Grade 7 1. Arranging the five numbers from largest to smallest, we get 10 000, 1000, 100, 10, 1. The middle number is 100. Answer: (D) 2. Each side of the square has length 5 cm. The perimeter of the square is 4 × 5 cm = 20 cm. Answer: (A) 3. The right side of the equation is 10 + 20 = 30. The equation is true when the left side is also equal to 30. Since 5 + 25 = 30, the value that goes in the box to make the equation true is 25. Answer: (E) 4. Reading from the graph, Dan spent 6 hours on homework, Joe spent 3 hours, Bob spent 5 hours, Susie spent 4 hours, and Grace spent 1 hour. Adding their times together, Bob and Grace spent the same amount of time on homework as Dan. Answer: (C) 5. Each of the five fractions is positive and so the smallest of these fractions is the fraction that is closest to 0. Since each fraction has a numerator equal to 1, the smallest of these fractions is the one with the largest denominator. Of those given, the fraction that is closest to 0 is thus 19. Answer: (E) 6. If the bag contained a total of 6 candies and exactly 5 of these candies were red, then the probability of Judith choosing a red candy from the bag would be 56. Therefore, the total number of candies in the bag could be 6. Can you explain why each of the other four answers is not possible? Answer: (D) 7. Each point that lies to the right of the y-axis has an x-coordinate that is positive. Each point that lies below the x-axis has a y-coordinate that is negative. Since P (x, y) lies to the right of the y-axis and below the x-axis, then the value of x is positive and the value of y is negative. Answer: (B) Andrew’s Walking Distance Distance (km) 8. Begin by locating 2 km on the vertical (Distance) axis. Next, locate the point on the line graph for which Andrew’s distance walked is 2 km, as shown. The time in hours corresponding to this point is 34 of an hour greater than 1 hour. Since 14 of an hour is equal to 15 minutes, then 34 of an hour is 45 minutes, and so it takes Andrew 1 hour, 45 minutes to walk the first 2

GaussCombinedContest-Sol.pdf

Document Details

Tags

Related

Full Transcript

Upgrade to continue