Chapter 5 Playing with Numbers PDF

## Properties of Square Numbers - All square numbers have the digit 0, 1, 4, 5, 6 or 9 in the units place. - Square numbers can only have an even number of zeros at the end. - The square root of an even number is even and the square root of an odd number is odd. ## Cubes **Example** - 6205 = 6...

## Properties of Square Numbers - All square numbers have the digit 0, 1, 4, 5, 6 or 9 in the units place. - Square numbers can only have an even number of zeros at the end. - The square root of an even number is even and the square root of an odd number is odd. ## Cubes **Example** - 6205 = 6 x 1000 + 2 x 100 + 5 x 1 - 918 = 9 x 100 + 1 x 10 + 8 x 1 - 176 = 1 x 100 + 7 x 10 + 6 - 460 = 4 x 100 + 6 x 10 + 0 - 719 = 7 x 100 + 1 x 10 + 9 - 36 = 3 x 10 + 6 - 147 = 1 x 100 + 4 x 10 + 7 x 1 ## Test of Divisibility - **Divisible by 2:** Last digit is 0, 2, 4, 6, 8 - **Divisible by 3:** If the sum of digits is divisible by 3, then the number is also divisible by 3. - **Divisible by 4:** If the last two digits are divisible by 4, then the number is also divisible by 4. - **Divisible by 5:** Last digit is 0 or 5. - **Divisible by 6:** Divisible by 2 and 3 - **Divisible by 8:** Last 3 digits are divisible by 8. - **Divisible by 9:** If sum of digits is divisible by 9, the number is divisible by 9. - **Divisible by 10:** Last digit is 0. - **Divisible by 11:** If the difference between the sum of the digits in its odd places and that of the digits in its even places is either 0 or divisible by 11, then the number is divisible by 11. ## Exercises **Exercise 1** - **39487362:** Digit is 2 hence it is divisible by 2. - **3+9+4+8+7+3+6+2 = 42** - 42 is divisible by 3, Hence given number is also divisible by 3. - **3+9+9+8+7+3+6+2 = 42** - 42 is not divisible by 9, Hence given number is not divisible by 9. **Exercise 2** - **37814626:** 3+7+8+1+4+6+2 = 31 - **31 is not divisible by 3 Hence the given number is not divisible by 3**. **Exercise 3** - 4 + 1 + 12 = 17 - 12 - 12 = 0 - **n = 8** - 3 + 7 + 10 = 20 - 19 - 10 = 9 - **n = 6** - 3 + 7 + 1 + 0 + 7 = 18 - **18 is divisible by 9, therefore the number is divisible by 9** **Exercise 4** - 6 + 2 + 0 + 7 = 15 - 15 is divisible by 3. Therefore the number is divisible by 3. - 5 + 1 + 2 + 2 + 8 = 18 - 18 is a complete multiple of 3. Hence the number is divisible by 3. - 1 + 5 + 1 + 9 + 5 = 21 - 21 is a complete multiple of 3. Hence the number is divisible by 3. - 2 + 3 + 1 + 5 + 5 + 2= 18 - Therefore, the number is divisible by 3. **Exercise 5** - 159472: The last digit is 2 and it divisible by 2. However, 1+5+9+4+7+2 = 28 is not divisible by 3 and hence not divisible by 6. - 556986: The last digit is 6 and it is divisible by 2. 5+5+6+9+8+6=39 is divisible by 3, thus, 556986 is divisible by 6 - 8234: The last digit is 4 and it is divisible by 2. However, 8+2+3+4 = 17 is not divisible by 3 and hence not divisible by 6. **Exercise 6** - 2560 - 6520 - 5620 - 2650 - 6250 - 6205 - 6025 - 5265 - 2065 - 2605 **Exercise 7** - The largest number you can write using only two digits with no mathematical symbols is **98** **Exercise 8** - (1)^3 = -1 - (-1)^2 = 1 **Exercise 9** - 23416 - 15624 - 12364 - 12456 - 13456 **Exercise 10** - 14625: 6 x 100 + 2 x 10 + 5 x 1 = 7817 - 341: 3 x 100 + 4 x 10 + 1 x 1 = 918 - 908: 9 x 100 + 0 x 10 + 8 x 1 = 4623 - 217: 2 x 100 + 1 x 10 + 7 x 1 = 89 **Exercise 11** - 3B6: A = 1, B = 9 - TA2C: A = 6, B = 9 - 519: A = 6, B = 3 - AGB: A = 9, B = 2 - 5C3: A = 5, B = 2 - 1045: A = 4, B = 5 **Exercise 12** - CILB: A = 5, B = 6 - 1239 A = 1, B = 5 - BAA: A = 5, B = 9 - A6: A = 1, B = 9 - XB: A = 2, B = 5 - A04: A = 3, B = 9 - B7: A = 1, B = 3 - XB: A = 2, B = 5 - AAA: A = 1, B = 3 **Exercise 13** - **The three digit numbers which are divisible by 37: ** - 456 - 465 - 546 - 564 - 645 - 1654 - 3330 **Exercise 14** - 6188 is not divisible by 9. - 2142 is divisible by 3 and 2, hence it is divisible by 6. - 12231 is divisible by 11. - 483952 is divisible by 4 because the last two digits are divisible by 4 - 41392 is not divisible by 9. - 7524 is divisible by 3, hence it is divisible by 3. **Exercise 15** - 420132: 4+2+0+1+3+2 = 12, therefore the number is divisible by 3. 4+2-0+1+3-2 = 8, therefore the number is divisible by 11. - 624083: 6+2+4+0+8+3 = 23, therefore the number is not divisible by 3. 6+2-4+0-8+3 = -1, therefore the number is not divisible by 11. **Exercise 16** - 44136: the last digit is 6, which is divisible by 2. 4+4+1+3+6 = 18, which is divisible by 3. Therefore, 44136 is divisible by 6. - 68324: the last digit is 4, which is divisible by 2. 6+8+3+2+4 = 23, which is not divisible by 3. Therefore, 68324 is not divisible by 6. - 218679: the last digit is 9, which is not divisible by 2. Therefore, 218679 is not divisible by 6. - 91428: the last digit is 8, which is divisible by 2. 9+1+4+2+8 = 24, which is divisible by 3. Therefore, 91428 is divisible by 6. - 5492736: the last digit is 6, which is divisible by 2. 5+4+9+2+7+3+6= 36, which is divisible by 3. Therefore, 5492736 is divisible by 6. **Exercise 17** - 12345678: The last digit is 8, it is divisible by 2. 1+2+3+4+5+6+7+8 = 36, it is divisible by 3. Hence the number is divisible by 6. **Exercise 18** - All possible three-digit numbers divisible by 3: - 405 - 450 - 504 - 540 - 409 - 490 - 904 - 940 - 509 - 590 - 905 - 950 - 495 - 459 - 545 - 945 - 954 - All possible three-digit numbers divisible by 9: - 540 - 405 - 450 - 945 - 954 - 549 - 594 - 504 - 495 - 459 ** Exercise 19** - By Pythagoras Theorem, (Hypotenuse)^2 = (Opposite)^2 + (Adjacent)^2. - In this case, Hypotenuse = 12, Opposite = 10, and Adjacent = 6. - 12^2 = 10^2 + 6^2 - 144 = 100 + 36 - 144 = 136 - The equation is not true. This is not a Pythagorean triplet. **Exercise 20** - There are 4 perfect square numbers between 100 and 200 which are: 100, 121, 144, and 169. - 89000 - 384 = 87,616 - 89000 - 1526 = 87,474 - 89000 - 7616 = 81,384 **Exercise 21** - 89000 can be multiplied by 2 to get 178000, which is a perfect square of 422. **Exercise 22** - The number of zeros in the square root of 729000000 is 3.