Playing with Numbers - Exercise 16.1

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Chapter 16: Playing with Numbers - Exercise 16.1

• Puzzles require each letter to stand for one digit only.

Question 1

• Problem: 3A + 25 = B2
• Difficulty: Easy
• Known: Addition operation of two numbers
• Unknown: Values of alphabets A and B
• Solution: A = 7 (since 7 + 5 yields a number ending in 2), which results in a carry-over of 1. The carry-over makes B = 6
• Therefore, A = 7 and B = 6

Question 2

• Problem: 4A + 98 = CB3
• Difficulty: Easy
• Known: Addition operation of two numbers
• Unknown: Values of alphabets A, B, and C
• Solution: A = 5 (since 5 + 8 yields a number ending in 3), which results in a carry-over of 1. Thus B = 4 and C = 1
• A = 5, B = 4 and C = 1

Question 3

• Problem: 1A x A = 9A
• Difficulty: Medium
• Known: Multiplication operation of two numbers
• Unknown: Value of alphabet A
• Solution: A = 6 (since 6 x 6 is results in a ones digit of 6), which results in a carry-over of 3. Thus 1x6=6 and adding the carryover results in 9
• A = 6

Question 4

• Problem: AB + 37 = 6A
• Difficulty: Medium
• Known: Addition operation of two numbers
• Unknown: Value of alphabets A and B
• Two cases
  • the first step isn't producing a carry over
    • this means that A=3, as 6-3=3. B should result in a ones digit of 3 after adding 7
    • this means B=6, as 6+7 = 13
    • this is impossible.
  • first step is producing a carry
    • in this case, A is 2 (1+3=6) and B has to be a number that would a unit's digit of 2 when 7 is added.
    • the solution is B=5 as 5+7=12.

Question 5

• Problem: AB x 3 = CAB
• Difficulty: Medium
• Known: Multiplication operation of two numbers
• Unknown: Value of alphabets A, B and C
• Solution: B has to be either 0 or 5, as B results in the ones digit after being multiplied by 3. After going through the process of elimination, B must be equal to 0
• A then must be equal to 5
• A = 5, B = 0 and C = 1

Question 6

• Problem: AB x 5 = CAB
• Difficulty: Medium
• Known: Multiplication operation of two numbers
• Unknown: Value of alphabets A, B and C
• Solution: B * 5 must give a number whose ones digit is B with that in mind, 5 or 0 would work.
• If B = 5, then B×5=5×5=25, 2 will be a carry-over. 5×A +2 = CA, which is possible if A=2 or 7

Question 7

• Problem: AB x 6 = BBBB
• Difficulty: Medium
• Known: Multiplication operation of two numbers
• Unknown: Value of alphabets A and B
• Solution: multiplication of 6 and B gives a number whose one's digit is B again giving B = 0, 2, 4, 6, or 8
• By process of elimination, B = 4, so, A = 7.

Question 8

• Problem: A1+1B=B0
• Difficulty: Medium
• Known: Addition operation of two numbers
• Unknown: Value of alphabets A and B
• Solution: The addition of 1 and B results in a one's digit of 0. B=9. Thus, in the next step, 1+A+1=B.
• A is 7

Question 9

• Problem: The addition 2AB+AB1=B18
• Difficulty: Medium
• Known: Addition of two numbers
• Unknown: Value of alphabets A and B
• Solution: the addition of B and 1 gives 8 giving digit B is 7. So, in the next step.
• A becomes 4.

Question 10

• Problem: The addition 12A+6AB=A09
• Difficulty: Medium
• Known: Addition operation of two numbers
• Unknown: Value of alphabets A and B
• Solution: Addition of A and B is giving 9 and two single digit numbers cannot be 1. A must be 8
• Thus B is 1

Chapter 16: Playing with Numbers - Exercise 16.2

• If the sum of all the digits of a number equals 9, the number is a multiple of 9.

Question 1

• Problem: If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
• Difficulty: Easy
• Known: A puzzled number
• Unknown: Value of the alphabet y
• Solution: Sum of digits of 21y5=2+1+y+5=8+y. If a number is a multiple of 9, then the sum of its digits will need to add up to 9 as well.
• Solution: y= 1

Question 2

• Problem: 31z5 is a multiple of 9, where z is a digit giving the value of z
• Difficulty: Easy
• Known: A puzzled number
• Unknown: Value of the alphabet y
• Solution: 9+z should be a multiple of 9
• Answer: z should be either 0 or 9

Question 3

• Problem: If 24x is a multiple of 3, find different values of x.
• Difficulty: Easy
• Known: A puzzled number
• Unknown: Value of the alphabet x
• Solution: The digits will need to have a multiple of 3 if 24x is a multiple of 3.
• x can have a value of 0,3,6 or 9

Question 4

• Problem: If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
• Difficulty: Easy
• Known: A puzzled number
• Unknown: Value of the alphabet z
• Solution: The sum of the digits must be divisible by 3. That is, 3+1+z+5=9+z is a multiple of 3.
• z can have its value as any one of the four different values 0, 3, 6, or 9.

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