Exercise practice 1: Number Play

3.1 Figure it Out, page 113 (first block)

3.1.1 FIO 1.1 Sum of four consecutive numbers

Question. Four consecutive integers add up to 34. Identify them.

Answer. 7, 8, 9 and 10.

Key steps. Let the numbers be n, n + 1, n + 2 and n + 3. Their sum is 4n + 6 = 34, so 4n = 28 and n = 7. The numbers are 7, 8, 9 and 10. Check: 7 + 8 + 9 + 10 = 34.

3.1.2 FIO 1.2 Five consecutive numbers

Question. Let p stand for the largest of five consecutive integers. Write the remaining four in terms of p.

Answer. p - 1, then p - 2, next p - 3, and finally p - 4.

Key steps. Counting backwards from the greatest number gives p - 1, then p - 2, p - 3 and p - 4. For example, if p = 12 the five numbers are 8, 9, 10, 11 and 12. The sum is 5p - 10, so the middle number is p - 2.

3.1.3 FIO 1.3 Always, sometimes or never

Question. Decide whether each claim holds in every case, in some cases only, or in no case, with examples and algebra: (i) adding two even numbers always gives a multiple of 3; (ii) if 18 does not divide a number, then 9 does not divide it either; (iii) if 6 divides neither of two numbers, it also does not divide their sum; (iv) Adding a multiple of 6 to a multiple of 9 always lands on a multiple of 3; (v) adding a multiple of 6 to a multiple of 3 always gives a multiple of 9.

Answer. (i) Sometimes true. (ii) Sometimes true. (iii) Sometimes true. (iv) Always true. (v) Sometimes true.

Key steps. (i) For the first claim, 2 + 4 = 6 works while 2 + 8 = 10 does not so the claim holds only sometimes. (ii) 27 is not divisible by 18 but is divisible by 9, while 30 is divisible by neither. So the implication fails sometimes and holds sometimes. (iii) For claim (iii), 6 divides neither 9 nor 11, and it also misses their sum 20, yet 8 and 10 also escape division by 6 while their sum 18 is divisible so the claim fails sometimes and holds sometimes. (iv) Write the numbers as 6x and 9y. Their sum is 6x + 9y = 3(2x + 3y), which always has 3 as a factor. (v) Adding 18 and 9 gives 27 which is a multiple of 9 whereas 12 + 9 = 21 is not. The sum 6x + 3y = 3(2x + y) is divisible by 9 only when 2x + y is divisible by 3.

3.1.4 FIO 1.4 Numbers with two remainders

Question. List several numbers giving remainder 2 on division by 3 and a remainder 2 on division by 4, then give an algebraic expression for all such numbers.

Answer. 14, 26, 38, ... and in general 12n + 2.

Key steps. Let the number be x. Then x - 2 is divisible by both 3 and 4, so x - 2 is a multiple of LCM(3, 4) = 12. Taking the smallest positive multiple x - 2 = 12 gives x = 14; the expression 12n + 2 also gives x = 2 when n = 0. For n = 1, 2 and 3 the values are 14, 26 and 38, and they continue the same pattern. Every number of the form 12n + 2 works because 12n is divisible by both 3 and 4, leaving remainder 2 in each case.

3.1.5 FIO 1.5 Pebble riddle

Question. A number of pebbles, fewer than 100: grouping by 3 leaves 1, grouping by 2 leaves 1, grouping by 5 leaves 1, and grouping by 7 leaves none. How many pebbles?

Answer. 91.

Key steps. Grouping by 3, by 2 and by 5 each leave remainder 1, so the number is 1 more than a common multiple of 3, 2 and 5. LCM(3, 2, 5) = 30, so the number is 30k + 1. Under 100 the possibilities are 31, 61 and 91. The number must also be divisible by 7: 31 and 61 are not, while 91 = 7 × 13. So there are 91 pebbles. The grouping by 2 leaving 1 is another way of saying the number is odd, which is why only the odd candidates 31, 61 and 91 need checking.

3.1.6 FIO 1.6 Tathagat's claim

Question. Tathagat writes several numbers that leave remainder 2 when divided by 6. He claims that the sum of any three such numbers is always a multiple of 6. Is the claim true?

Answer. Yes, it is always true.

Key steps. Write the three numbers as 6a + 2, 6b + 2 and 6c + 2. Their sum is 6(a + b + c) + 6 = 6(a + b + c + 1), which has 6 as a factor. The remainders 2 + 2 + 2 = 6 combine into one extra group of 6, so the total is a multiple of 6.

3.1.7 FIO 1.7 Remainders of sums and differences

Question. When divided by 7, 4779 leaves remainder 5 and 661 leaves remainder 3. Without calculating the numbers, find the remainders of (i) 4779 + 661 and (ii) 4779 - 661 when divided by 7.

Answer. (i) 1. (ii) 2.

Key steps. Write 4779 = 7p + 5 and 661 = 7q + 3. (i) Sum = 7p + 7q + 8 = 7(p + q) + 8 = 7(p + q + 1) + 1 so the remainder is 1. (ii) Difference = 7p + 5 + 7q + 3 = 7(p + q) + 8 = 7(p + q + 2) so the remainder is 2. In both cases the non-multiple part is combined separately, and the leftover is the remainder.

3.1.8 FIO 1.8 Three remainders, one less each

Question. Find a number that leaves remainder 2 when divided by 3, remainder 3 when divided by 4 and remainder 4 on division by 5. Find the least such number.

Answer. 59.

Key steps. Each remainder is one less than its divisor: 2 = 3 - 1, 3 = 4 - 1 and 4 = 5 - 1. So the number plus 1 is divisible by 3, 4 and 5. The smallest number divisible by all three is LCM(3, 4, 5) = 60. Therefore the smallest number with the required remainders is 60 - 1 = 59. It is the smallest because any smaller number plus 1 would be a smaller common multiple of 3, 4 and 5, and 60 is the least one.

3.2 Figure it Out, page 126 (second block)

3.2.1 FIO 2.1 Divisibility by 9

Question. Without dividing, decide whether each number below is divisible by 9: 123, 405, 8888, 93547 and 358095.

Answer. Only 405 is divisible by 9.

Key steps. Add the digits of each number. (i) 1 + 2 + 3 = 6, not divisible by 9. (ii) 4 + 0 + 5 = 9, divisible by 9, so 405 is divisible. (iii) 8 + 8 + 8 + 8 = 32, not divisible by 9. (iv) 9 + 3 + 5 + 4 + 7 = 28, not divisible by 9. (v) 3 + 5 + 8 + 0 + 9 + 5 = 30, not divisible by 9.

3.2.2 FIO 2.2 Least multiple of 9 whose decimal digits are all even

Question. Find the least multiple of 9 whose decimal digits are all even.

Answer. 288.

Key steps. The allowed digits are 2, 4, 6 and 8. Their sums are even, so the digit sum can never be 9; the smallest possible digit sum that is a multiple of 9 is 18. To keep the number small, use as few digits as possible and the smallest leading digit. The combination of allowed digits with sum 18 is 2 + 8 + 8. Arranging them in ascending order gives 288, and 288 = 9 × 32. No one-digit or two-digit number with all even digits has digit sum 18 (the largest two-digit sum is 8 + 8 = 16), so 288 is the smallest.

3.2.3 FIO 2.3 Multiple of 9 closest to 6000

Question. Find the multiple of 9 closest to 6000.

Answer. 6003.

Key steps. Check the digit sum: 6 + 0 + 0 + 0 = 6, so 6000 is not divisible by 9 and leaves remainder 6. The next multiple below is 6000 - 6 = 5994, with digit sum 5 + 9 + 9 + 4 = 27. The next multiple above is 6000 + 3 = 6003, with digit sum 9. Since 3 < 6, 6003 is closer than 5994.

3.2.4 FIO 2.4 Multiples of 9 between 4300 and 4400

Question. How many multiples of 9 lie between 4300 and 4400?

Answer. 11.

Key steps. The first multiple of 9 after 4300: the digit sum of 4300 is 7, so add 2 to reach 4302, whose digit sum is 9. The last multiple before 4400: 4400 has digit sum 8, so subtract 8 to get 4392, whose digit sum is 18. The multiples form the sequence 4302, 4311, 4320, ... 4392 with a common difference of 9. The count is (4392 - 4302)/9 + 1 = 90/9 + 1 = 11.