AMC 10 · 2018 · #13

Grade 6 number-theory
Archetype Modular Cycle Frequency Count └ Sub-archetype Floor Count Across Residue Bins
modular-arithmeticplace-valuepattern-recognition pattern-recognitioneasier-related-problem ↑ Prerequisites: modular-arithmetic
📏 Medium solution 💡 2 insights
Problem
Look at the list 101, 1001, 10001, 100001, …, where each number is a 1, then a block of zeros, then another 1. Count how many of the first 2018 numbers in this list are exact multiples of 101.

Pick an answer.

(A)
253
(B)
504
(C)
505
(D)
506
(E)
1009

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Tool #5 (Look for a Pattern): each term is a power of ten plus one, and remainders of powers of ten when divided by 101 repeat in a short cycle. Finding that cycle turns a question about huge numbers into a question about a repeating list of four. Tool #9 (Solve an Easier Related Problem): instead of testing each giant number, we replace 'is the term divisible by 101' with the much smaller question 'what remainder does a power of ten leave when divided by 101'. Tool #2 (Make a Systematic List): once we know which positions work, the answer is just counting the positions that fit a simple rule up to 2018.

1STEP 1

Write each term with a power of ten

Each term is a 1, some zeros, then a 1 — that is a power of ten plus one, with exponent one bigger than the position.

a_n = 10^ n+1 + 1
2STEP 2

Turn divisibility into a remainder

The term is a multiple of 101 exactly when the power of ten before the +1 leaves remainder 100 mod 101.

101 ∣ 10^ n+1+1 ⇔ 10^ n+1 leaves remainder 100 (mod 101)
3STEP 3

Find the cycle of remainders

Powers of ten mod 101 give remainders 10, 100, 91, 1, then repeat — a cycle of length 4.

10¹,10²,10³,10⁴ → 10, 100, 91, 1, 10, 100, 91, 1,…
4STEP 4

Pick out the winning positions

Only the second cycle slot gives remainder 100, so the working positions are n = 1, 5, 9, 13, …

10^ n+1≡ 100 ⇔ n+1≡ 2 (mod 4) ⇔ n = 1,5,9,13,…
5STEP 5

Count those positions up to 2018

The positions 1, 5, 9, …, 2017 are n = 1+4k for k = 0…504, giving 505 terms — choice (C).

n = 1+4k ≤ 2018 → k ≤ 504 → 505 values → (C)
Answer
505
About one position in every four works, and 2018÷4≈504.5, so the answer should sit right around 504 or 505 — which rules out the far-off choices (A) 253 and (E) 1009. The exact value lands on 505, not 504, because the very first term (n=1, the number 101 itself) already counts, nudging the tally up by one. The off-by-one traps (B) 504 and (D) 506 come from miscounting that boundary; carefully, n=1 through n=2017 in steps of 4 gives exactly 505.
💡Key takeaway

Remainders of powers of ten by 101 loop every four steps, so only every fourth number in the list is divisible by 101 — and counting those up to position 2018 gives 505.

  • Write each term with a power of ten
  • Turn divisibility into a remainder
  • Find the cycle of remainders
  • Pick out the winning positions
  • Count those positions up to 2018

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