Competition · AMC preparation · step 4 of 4

AMC 10 · 2011B · #23

Grade 8 arithmetic
Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue
modular-arithmeticbinomial-theoremplace-value easier-related-problem ↑ Prerequisites: modular-arithmetic
📏 Medium solution 💡 3 insights
Problem
Find the hundreds digit of the giant number 2011²⁰¹¹ — the digit in the third place from the right.

Pick an answer.

(A)
1
(B)
4
(C)
5
(D)
6
(E)
9

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

The number 2011²⁰¹¹ is astronomically large, but the question asks for a single digit near the right end. Tool #9 (Solve an Easier Related Problem) collapses the giant into something small: the hundreds digit is fixed by the last three digits, and last three digits of a product depend only on the last three digits of the factors — so 2011²⁰¹¹ and 11²⁰¹¹ have the same last three digits. Now we only need 11²⁰¹¹ mod 1000. Tool #7 (Identify Subproblems) cracks that open: write 11 = 10 + 1 and expand (10+1)²⁰¹¹; almost every term is a multiple of 1000 and vanishes, leaving just three pieces to compute. Tool #2 (Make a Systematic List) adds those three pieces. Tool #5 (Look for a Pattern) gives an independent cross-check.

1STEP 1

Shrink the giant to its last three digits

Carries only travel leftward, so since 2011 ends in 011, 2011²⁰¹¹ and 11²⁰¹¹ share the same last three digits.

2011²⁰¹¹ ≡ 11²⁰¹¹ (mod 1000)
2STEP 2

Rewrite 11 as 10 + 1

Expand (10+1)²⁰¹¹: every term using three or more 10's carries 10³ = 1000 and vanishes, so only zero, one, or two 10's survive.

11²⁰¹¹ = (10+1)²⁰¹¹, 10³ = 1000 ≡ 0 (mod 1000)
3STEP 3

Keep only the three surviving pieces

Survivors: 1 from no tens; 2011 × 10 = 20110 gives 110; (2011×2010)/2 = 2021055 hundreds gives 500.

1 + 2011×10_≡ 110 + (2011×2010)/2×100_≡ 500 (mod 1000)
4STEP 4

Add the pieces and read the digit

Add them mod 1000: 1 + 110 + 500 = 611, so 2011²⁰¹¹ ends in 611 and the hundreds digit is 6 — choice (D).

11²⁰¹¹ ≡ 1 + 110 + 500 = 611 (mod 1000) → hundreds digit = 6 → (D)
5STEP 5

Cross-check with an independent method

Repeated squaring mod 1000 agrees: 11² ≡ 121, 11⁴ ≡ 641, 11⁸ ≡ 881, and recombining those powers lands on 611 again.

11²⁰¹¹ ≡ 11 (mod 100) and repeated squaring → 611 (mod 1000)
Answer
6
The last three digits 611 end in 11, exactly as expected since any power of a number ending in 11 still ends in …1 in the units and picks up 11 in the last two places here. Two independent computations — the truncated (10+1)²⁰¹¹ expansion and repeated squaring modulo 1000 — both give 611, so the hundreds digit 6 is solid. Among the choices, only (D) 6 matches; the distractors 1, 4, 5, 9 correspond to common slips (dropping the two-tens term, mis-halving 2011 × 2010, or reading the wrong place).
💡Key takeaway

For the hundreds digit of 2011²⁰¹¹, throw away everything but the last three digits: 2011 becomes 11, expanding (10+1)²⁰¹¹ leaves only three pieces, and 1 + 110 + 500 = 611 — so the hundreds digit is 6.

  • Shrink the giant to its last three digits
  • Rewrite 11 as 10 + 1
  • Keep only the three surviving pieces
  • Add the pieces and read the digit
  • Cross-check with an independent method

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