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2011 AMC 8

Problem 22

Problem 22 · 2011 AMC 8 Hard
Number Theory tens-digit-cyclemod-100

What is the tens digit of 72011?

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Answer: D — Tens digit 4.
Show hints
Hint 1 of 2
Computing 72011 is hopeless — but the tens digit only depends on the last two digits, and those depend only on the last two digits of the previous power. So track just the last two digits as you multiply by 7.
Still stuck? Show hint 2 →
Hint 2 of 2
List the last-two-digit patterns: 07, 49, 43, 01, then it returns to 07. A repeating cycle of length 4 — now the only question is where 2011 lands in the cycle.
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Approach: only the last two digits matter, and they cycle with period 4
  1. Multiply keeping just the last two digits: 7, then 49, then 7×49 = 343 → 43, then 7×43 = 301 → 01, then 7×01 = 07 — back to the start. The cycle 07, 49, 43, 01 repeats every 4 powers.
  2. Find 2011's spot: 2011 = 4·502 + 3, a remainder of 3, so 72011 ends like the 3rd entry, 73 = 43.
  3. The tens digit of …43 is 4.
  4. Worth keeping: to get the last k digits of a big power, work ‘mod 10k’ (here mod 100) — last digits always settle into a short repeating cycle, so you only need the exponent's remainder.
Another way — shortcut via 74 ≡ 1 (mod 100):
  1. 74 = 2401 ends in 01, i.e. 74 ≡ 1 (mod 100), so any 74 block can be dropped.
  2. 2011 = 4·502 + 3, so 72011 ≡ (74)502·73 ≡ 1·43 = 43 (mod 100); tens digit 4.
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