AMC 10 · 2005 · #11
Grade 6 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Computing 2005 terms directly is impossible, so Tool #5 (Look for a Pattern) is the main move: generate the first several terms and watch for a value that repeats, which forces the whole sequence into a cycle. Tool #7 (Identify Subproblems) handles each single step — one cube-of-digits calculation at a time — so the arithmetic stays clean. Tool #9 (Solve an Easier Related Problem) is what finishes it: instead of reaching term 2005 the long way, we replace that with an easy division-remainder question about where 2005 lands inside the short repeating cycle.
Compute the second term
Cube each digit of 2005 and add: 2³+0³+0³+5³ = 8+125 = 133, the second term.
Cubing a digit just means multiplying it by itself three times, and the zeros drop out because 0³=0.
6.EE.A.1Identify SubproblemsCompute the next few terms
Repeat: 133 → 1+27+27 = 55, then 55 → 125+125 = 250, then 250 → 8+125+0 = 133 again.
Each arrow is the same rule applied once, so you only ever need one small cube-and-add calculation at a time.
6.EE.A.1Identify SubproblemsSpot the repeating cycle
Term 5 repeats term 2, and each term depends only on the one before, so 133, 55, 250 cycles with period 3.
A rule that only looks at the previous term must loop forever the moment any value shows up a second time.
A rule that only looks at the previous term must loop forever once any value shows up twice.
▸ Why?
Once a value repeats, everything that followed it the first time follows it again in the same order.
▸ Why?
Inside a loop of fixed length only the remainder after dividing decides where a step lands.
Find where the 2005th term lands
Term 4 has remainder 1 mod 3 and equals 250; since 2005 = 3×668+1 also gives remainder 1, the answer is 250 = (E).
In a length-3 loop, only the remainder after dividing by 3 decides which of the three values you land on.
4.NBT.B.6Solve An Easier Related ProblemWhen a rule keeps feeding a term back into itself, list a few terms until a value repeats, then use the remainder after dividing by the loop length to jump straight to any far-off term.
- Compute the second term
- Compute the next few terms
- Spot the repeating cycle
- Find where the 2005th term lands