AMC 10 · 2021 · #25

Grade 6 number-theory
modular-arithmeticdivisibility-rulessystematic-enumerationcasework systematic-enumerationcasework ↑ Prerequisites: modular-arithmeticdivisibility-rules
📏 Long solution 💡 4 insights
Problem
For a positive integer, divide it by each of 2 through 10 and add up the nine remainders. Among the two-digit values, count how many give the same total as the next number does.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4
How to solve
Strategy Change Focus / Count the Complement

Computing R(n) and R(n+1) and comparing them is the obvious move, and it is the wrong one — it does twice the work and throws away the fact that the two inputs differ by exactly 1. Tool #16 (Change Focus) says to stop tracking the totals and track only the change R(n+1) - R(n). That change splits into nine independent pieces, one per divisor, and tool #5 (Look for a Pattern) settles each piece with a single observation: adding 1 pushes a remainder up by 1, unless it has just hit a multiple, in which case it falls all the way to 0. Tool #13 (Convert to Algebra) then adds the nine pieces into one clean equation whose only ingredient is the list of numbers from 2 to 10 that divide n+1. From there the problem is arithmetic on small sets: tool #2 (Make a Systematic List) writes down every subset of {2,3,…,10} whose sum is 9, tool #3 (Eliminate Possibilities) throws out the ones that no number could actually have as its divisor list, and tool #6 (Guess and Check) screens the handful of survivors in the two-digit range.

1STEP 1

Track the change, not the totals

Track the change, not the totals.

D(n) = R(n+1) - R(n) = Σ_k=2¹⁰ [((n+1) mod k) - (n mod k)]; the goal is D(n) = 0
2STEP 2

What adding one does to a remainder

Adding one changes a remainder in one of two ways.

((n+1) mod k) - (n mod k) = +1 & if k ∤ n+1 ; -(k-1) & if k ∣ n+1
3STEP 3

Add the nine changes together

Add the nine changes together.

D(n) = 9 - S(m), m = n+1, S(m) = Σ_{2 ≤ k ≤ 10 ; k ∣ m} k; R(n) = R(n+1) ⇔ S(m) = 9
4STEP 4

List every subset summing to nine

List every subset summing to nine.

{9}, {2,7}, {3,6}, {4,5}, {2,3,4} — all subsets of {2,…,10} with sum 9
5STEP 5

Most of those sets are impossible

Most of those sets are impossible.

9 ∣ m → 3 ∣ m; 6 ∣ m → 2 ∣ m; 4 ∣ m → 2 ∣ m; lcm(3,4) = 12 → 6 ∣ m. Survivor: {2,7}, i.e. 14 ∣ m
6STEP 6

Screen the multiples of fourteen

Screening the multiples leaves 2.

m ∈ {14, 28, 42, 56, 70, 84, 98}; discard 28, 56 (divisible by 4), 42, 84 (by 3), 70 (by 5) → m = 14, 98 → n = 13, 97
Answer
2
Both winners can be checked against the original definition with no theory at all. For n = 13 the remainders are 1, 1, 1, 3, 1, 6, 5, 4, 3, summing to 25; for n = 14 they are 0, 2, 2, 4, 2, 0, 6, 5, 4, also summing to 25. For n = 97 they are 1, 1, 1, 2, 1, 6, 1, 7, 7, summing to 27; for n = 98 they are 0, 2, 2, 3, 2, 0, 2, 8, 8, again 27. So two values genuinely work, which already rules out (A) and (B). The count cannot be larger, because step 5 left {2,7} as the only possible divisor list and step 6 exhausted the multiples of 14 below 101 — so (D) and (E) are closed off too. It is also easy to see where a wrong count would come from: keeping {2,3,4} without noticing that 12 ∣ m forces 6 ∣ m would send a solver hunting among multiples of 12, and keeping {4,5} would send them to multiples of 20, each inflating the total. One more consistency note: 98 = 7 · 14, so both survivors are 14 times an odd number free of 3 and 5, and the next such multiple, 14 · 11 = 154, is already past two digits — which is exactly why the count stops at two.
💡Key takeaway

When a question compares a quantity at n with the same quantity at n+1, do not compute both — compute only what changes, because adding 1 nudges every remainder up by one except at the multiples, where it crashes to zero.

  • Track the change, not the totals
  • What adding one does to a remainder
  • Add the nine changes together
  • List every subset summing to nine
  • Most of those sets are impossible
  • Screen the multiples of fourteen