Mixed practice: Number Play

This page is mixed practice for the whole chapter. The questions are original and combine divisibility tests, digital roots, consecutive numbers and digit patterns. Work each one without a calculator, using the algebra of place value wherever a pattern is claimed. Write the rule you use, then the calculation, then a one-line check. A correct number without the rule earns few marks, so show the test and the digits you used.

5.1 Task 1: Several divisibility tests at once

Without dividing, decide whether each of the numbers 2, 3, 4, 5, 6, 8, 9 and 11 divides 3,45,672.

Solution. 2: the last digit is 2, so yes. 3: the digit sum is 3 + 4 + 5 + 6 + 7 + 2 = 27, and 27 is divisible by 3, so yes. 4: the last two digits are 72, and 72 is divisible by 4, so yes. 5: the last digit is not 0 or 5, so no. 6: the number is divisible by 2 and by 3, so yes. 8: the last three digits are 672, and 672 = 8 × 84, so yes. 9: the digit sum is 27, which is divisible by 9, so yes. 11: (3 + 5 + 7) - (4 + 6 + 2) = 15 - 12 = 3, which is not a multiple of 11, so no.

5.2 Task 2: Digital roots and remainders

Find the digital root of 4,78,652. Use it to find the remainder the number leaves on division by 9 and by 3, and find the least number to add so that the result is a multiple of 9.

Solution. The digit sum is 4 + 7 + 8 + 6 + 5 + 2 = 32, and 3 + 2 = 5, so the digital root is 5. A number and its digit sum leave the same remainder on division by 9, so the remainder is 5. For division by 3 the digit sum 32 leaves remainder 2. To reach the next multiple of 9, add 9 - 5 = 4: the number becomes 4,78,656, whose digit sum is 36 and whose digital root is 9.

5.3 Task 3: Consecutive numbers

(a) Prove that the sum of any three consecutive integers is a multiple of 3. (b) Five consecutive integers add to 95. Find them.

Solution. (a) Write the three integers as n - 1, n and n + 1. Their sum is 3n, which is a multiple of 3 for every whole number n. (b) The sum of five consecutive integers is five times the middle one, so the middle number is 95/5 = 19. The numbers are 17, 18, 19, 20 and 21, and their sum is 95.

5.4 Task 4: Products of consecutive numbers

Prove that the product of any three consecutive integers is divisible by 6.

Solution. Among three consecutive integers one is a multiple of 3, because every third integer is a multiple of 3. At least one of the three is even, so the product is also a multiple of 2. Since the product is a multiple of both 2 and 3, and 2 and 3 are coprime, it is a multiple of 6. Check with 7, 8 and 9: 7 × 8 × 9 = 504 = 6 × 84.

5.5 Task 5: The missing digit

The four-digit number 74d6 is divisible by 9. Find d, then test the resulting number for divisibility by 4 and by 11.

Solution. The digit sum is 7 + 4 + d + 6 = 17 + d. For divisibility by 9 this sum must be a multiple of 9; between 17 and 26 the only multiple of 9 is 18, so d = 1 and the number is 7416. Divisibility by 4: the last two digits are 16, so yes. Divisibility by 11: (7 + 1) - (4 + 6) = 8 - 10 = -2, which is not a multiple of 11, so no.

5.6 Task 6: Two tests together

Find every digit z for which the number 4z72 is divisible by both 3 and 4.

Solution. Divisibility by 4 depends only on the last two digits, which are 72, so it holds for every z. Divisibility by 3 needs the digit sum 4 + z + 7 + 2 = 13 + z to be a multiple of 3. The possible sums are 15, 18 and 21, so z = 2, 5 or 8. The numbers are 4272, 4572 and 4872, and their digit sums are 15, 18 and 21 respectively.

5.7 Task 7: A reversal pattern

Prove that a two-digit number plus its reversal is always a multiple of 11. Then find all two-digit numbers whose reversal is also two-digit and whose sum with their reversal is 99.

Solution. Write the number as 10a + b, where a is from 1 to 9. Its reversal is 10b + a. The sum is 11a + 11b = 11(a + b), which is a multiple of 11. For the sum to be 99, a + b = 9. With both digits non-zero, the numbers are 18, 27, 36, 45, 54, 63, 72 and 81.