AMC 8 · 2011 · #22
Grade 6 number-theoryPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The exponent 2011 is enormous, so direct computation is hopeless. Tool #9 (Easier Problem) says: try smaller exponents first — 7¹, 7², 7³, … — and only keep the last two digits at each step (anything earlier than the tens place can never affect the tens digit). Tool #5 (Look for a Pattern) takes over once those last-two-digit values start repeating: a short cycle lets us replace exponent 2011 with a much smaller equivalent exponent.
Keep only the last two digits as you raise 7 step by step: 07, 49, 43, 01 — anything past the tens place can never reach it.
Replacing a huge exponent with 1, 2, 3, 4 is the Easier Problem move. Tracking only the last two digits uses the Grade 5 place-value idea that the tens and ones come from below the hundreds.
5.NBT.A.1Solve An Easier Related ProblemMultiply once more: 01 × 7 = 07, right back to the start — so the last two digits repeat with period 4.
Detecting and stating a repeating cycle is the Grade 4 "generate and analyze patterns" skill — once 7⁴ returns to 01, the cycle must restart.
4.OA.C.5Look For A PatternDivide the exponent by the cycle length: 2011 = 4 × 502 + 3, giving remainder 3 — so 7²⁰¹¹ ends like 7³.
Using division-with-remainder to locate a number inside a repeating cycle is the Grade 6 "divide multi-digit numbers" skill applied to pattern position.
6.NS.B.2Look For A PatternSince 7³ ends in 43, so does 7²⁰¹¹ — the tens digit is 4, choice (D).
Identifying which digit sits in the tens place is the Grade 5 place-value definition itself.
5.NBT.A.1Look For A PatternHuge exponents look scary, but the last two digits cycle quickly — a Grade 6 division with remainder is all you need to land the answer!