AMC 10 · 2009 · #21
Grade 4 number-theoryPick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The sum is far too big to add up, so Tool #9 (Solve an Easier Related Problem) shrinks it: the remainder of a sum divided by 8 depends only on the remainders of the parts, so replace each power 3^k by its small remainder. Tool #5 (Look for a Pattern) is the engine: the remainders of 3⁰,3¹,3²,… do not grow, they loop in a short cycle, so we never touch the huge numbers. Tool #7 (Identify Subproblems) turns "add 2010 remainders" into the easy sub-task of pairing them, and Tool #3 (Eliminate Possibilities) matches the final remainder to one of the five listed choices.
Find the remainder cycle
Divided by 8, the powers of 3 leave 1,3,1,3,…: since 3²=9 is back to 1, the cycle has length 2 — even exponents leave 1, odd leave 3.
Once a remainder repeats, the whole list loops, so you only need the short cycle, not the giant powers.
Once a remainder repeats, the whole list loops, so only the short cycle is needed.
▸ Why?
The next remainder depends only on the current one, so a repeat forces the same run to follow.
▸ Why?
Inside a loop of fixed length only the leftover after dividing decides where a term lands.
Swap each power for its remainder
A sum's remainder is fixed by its pieces' remainders, so replace the giant powers by 1+3+1+3+…, one small entry for each of the 2010 terms.
Remainders add just like the numbers do, so the messy sum and the tidy remainder sum leave the same leftover.
3.OA.B.5Solve An Easier Related ProblemPair them and total
Group them two at a time: 2010 terms make 1005 pairs of (1+3), each worth 4, so the remainder sum is 1005×4 = 4020.
Bundling the loop into equal pairs turns a long addition into one multiplication.
4.NBT.B.5Identify SubproblemsDivide and match the choice
Divide the tidy total: 4020 = 8×502 + 4, so the original giant sum leaves remainder 4 too — choice (D).
The remainder of the small stand-in total is the remainder of the real sum, so one short division finishes it.
4.NBT.B.6Eliminate PossibilitiesPowers of 3 leave remainders that loop 1,3,1,3 when divided by 8; pair them up, and 1005 pairs of 4 leave remainder 4 — choice (D).
- Find the remainder cycle
- Swap each power for its remainder
- Pair them and total
- Divide and match the choice