Exercise practice 2: Number Play

4.1 Figure it Out, page 131 (third block)

4.1.1 FIO 3.1 Digital root after adding 10

Question. An eight-digit number has digital root 5. What is its digital root after adding ten?

Answer. 6.

Key steps. Adding 10 increases the digit sum by 1 in every case except when carrying changes the digit count. A number with digital root 5 has digit sum 5 modulo 9. Adding 10 adds 1 to the digit sum modulo 9, so the new digital root is 5 + 1 = 6. For example, 40,000,001 has digital root 5, and 40,000,011 has digit sum 4 + 1 + 1 = 6.

4.1.2 FIO 3.2 Digital roots in a sequence

Question. Start with any number and repeatedly add 11. What are the digital roots of the sequence? Share your observations.

Answer. Starting from 10, the sequence is 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, ... and its digital roots are 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, 3, ...

Key steps. Adding 11 changes the remainder modulo 9 by 11 mod 9 = 2. So each digital root is 2 more than the previous one, with 9 followed by 2 because 10 has digital root 1. The digital roots cycle through 1, 3, 5, 7, 9, 2, 4, 6, 8 and repeat. Every ninth step returns to the same digital root, so the sequence is periodic.

4.1.3 FIO 3.3 Digital root of an expression

Question. What is the digital root of 9a + 36b + 13, where a and b are whole numbers?

Answer. 4.

Key steps. 9a and 36b are multiples of 9, so they leave remainder 0 on division by 9 and do not change the remainder. The digital root depends only on the remaining term: 13 has digit sum 1 + 3 = 4, so the digital root is 4. For example, with a = 1 and b = 0, 9 + 13 = 22, whose digital root is 4; with a = 2 and b = 1, 18 + 36 + 13 = 67, whose digital root is 4.

4.1.4 FIO 3.4 Conjectures about digital roots

Question. Examine patterns between (i) a number's parity and its digital root, and (ii) the digital root and the remainder when that number is divided by 3 or 9.

Answer. (i) There is no fixed relation between the parity of a number and the parity of its digital root. For example, 14 is even and has digital root 5, which is odd; 20 is even and has digital root 2, which is even. The difference between a number and its digital root is a multiple of 9, and a multiple of 9 can be odd or even, so parity can change. (ii) The digital root gives the remainder modulo 9: a digital root of 1 to 8 is exactly the remainder on division by 9, and a digital root of 9 means the remainder is 0. For division by 3, take the digital root modulo 3: digital roots 3, 6 and 9 give remainder 0, while the others give the same remainder as the digital root.

Key steps. Write the number as N = 9q + r. The digital root is r when r is not 0, and 9 when r = 0. Since 9q is divisible by 9 and also by 3, the remainder modulo 3 is r mod 3 unless r = 9, in which case it is 0. The parity example shows why a relation with parity does not hold: N - dr = 9q, and 9q has the parity of q, which is not fixed.

4.2 Figure it Out, page 132 (fourth block)

4.2.1 FIO 4.1 The digit z in 31z5

Question. 31z5 is divisible by 9 and z stands for a digit. Find z and explain why there are two answers.

Answer. z = 0 or z = 9.

Key steps. The digit sum is 3 + 1 + z + 5 = 9 + z. For the number to be divisible by 9, 9 + z must be a multiple of 9. With z a digit from 0 to 9, the possible sums are 9 (z = 0) and 18 (z = 9). Both work: 3105 and 3195 are divisible by 9. There are two answers because a digit sum can be 9 or 18 and still be a multiple of 9.

4.2.2 FIO 4.2 Snehal's claim

Question. Snehal adds a number that leaves remainder 8 on division by 12 to a number lying 4 below a multiple of 12. She says the result is always a multiple of 8. Is she right?

Answer. No, the claim is false.

Key steps. Let the first number be 12n + 8 and the second be 12m - 4. Their sum is 12(n + m) + 4 = 12k + 4. This is not always a multiple of 8. Take k = 2: 12 × 2 + 4 = 28, and 28 is not divisible by 8. A single counterexample disproves the "always" claim; the sum is a multiple of 8 only for some values of k, for example k = 1 gives 16.

4.2.3 FIO 4.3 Sum of two multiples of 3

Question. Under what condition does adding two multiples of 3 produce a result divisible by 6 and when is it not? Generalise.

Answer. Let the multiples be 3m and 3n. Their sum is 3(m + n). It is a multiple of 6 exactly when m + n is even, that is, when m and n have the same parity.

Key steps. If both multiples are even multiples of 3 (such as 6 and 12), then m and n are both even, so m + n is even. If both are odd multiples of 3 (such as 3 and 9), then m and n are both odd, so m + n is even too. If one is an even multiple and the other an odd multiple (6 and 3), then m + n is odd and the sum 9 is not a multiple of 6. Hence the cases split by the parity of the multipliers, not by the multiples themselves.

4.2.4 FIO 4.4 Reversing the digits of a multiple of 9

Question. Sreelatha says a number divisible by 9 remains divisible by 9 when its digits are reversed. (i) Is the conjecture true for every multiple of 9 (ii) are other digit shuffles possible that also keep the number a multiple of 9?

Answer. (i) Yes. (ii) Any permutation of the digits also gives a multiple of 9 (provided the new number does not start with 0).

Key steps. Divisibility by 9 depends on the sum of the digits. Reversing the digits or rearranging them does not change which digits are present, so the digit sum is unchanged. If the original sum is divisible by 9, the sum of the shuffled digits is also divisible by 9, so the new number is divisible by 9. For example, 108 is divisible by 9; its permutations 180, 801 and 810 are also divisible by 9. A permutation such as 018 is not written as a three-digit number because a leading zero is dropped, but as 18 it is still divisible by 9.

4.2.5 FIO 4.5 The number 48a23b

Question. 48a23b is divisible by 18 find every possible pair (a, b).

Answer. (a, b) = (1, 0), (2, 8), (4, 6), (6, 4) and (8, 2).

Key steps. A multiple of 18 must be divisible by 2 and by 9. Divisible by 2 means b is even: b ∈ {0, 2, 4, 6, 8}. Divisible by 9 means the digit sum is a multiple of 9: adding 4 8 a 2 3 and b gives 17 a b. The smallest possible sum is 17 and the largest is 35, so the sum must be 18 or 27. Hence a + b = 1 or a + b = 10.

For a + b = 1 with b even, the only possibility is b = 0 and a = 1. For a + b = 10 with b even, the pairs are (a, b) = (2, 8), (4, 6), (6, 4) and (8, 2); b = 0 would give a = 10, which is not a digit. These five pairs all give a number divisible by 18.