AMC 10 · 2005 · #10
Grade 6 arithmeticPick an answer.
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
The second term comes out as 133.
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
Continuing gives 55, then 250, then 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
Since each term depends only on the last, the sequence cycles with period three.
A rule that only looks at the previous term must loop forever the moment any value shows up a second time.
A rule that looks only at the previous term must loop forever the moment any value shows up a second time.
▸ 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 by that length decides where a step lands.
Find where the 2005th term lands
Matching the remainder to the offset gives 250, choice (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