2007 AMC 10A Problem 22

Picture the terms as a ring of digits $d_1, d_2, \ldots, d_k$ read cyclically: the $i$-th term is $d_i d_{i+1} d_{i+2}$ (indices wrapping around). Each digit $d_i$ serves as the hundreds digit of one term, the tens digit of the previous term, and the units digit of the one before that.

So if $T = d_1 + d_2 + \cdots + d_k$ is the total of the ring's digits, then
$$S = 100T + 10T + T = 111\,T = 3 \cdot 37 \cdot T.$$
Thus $37$ always divides $S$.

No larger prime is guaranteed: the sequence $123, 231, 312$ is valid and sums to $666 = 2 \cdot 3^2 \cdot 37$, which is not divisible by $43$ (or by any prime above $37$).

The answer is $\boxed{\textbf{(D)}\ 37}$.

The wrap-around condition makes the multiset of hundreds digits, tens digits and units digits identical, so the sum factors as $111$ times something. "Always divides" questions have two halves: a structural argument that the divisor works for every case, and one explicit example showing nothing bigger does. Recognize $111 = 3 \cdot 37$ (and $1001 = 7 \cdot 11 \cdot 13$) on sight.

Complete the given example into a valid cycle, such as $247, 475, 756, 568, 682, 824$, and add: $S = 3552 = 2^5 \cdot 3 \cdot 37$. Only $3$ and $37$ among the choices divide it, and $37$ is larger; the ring argument confirms $37$ works in general.

• Concluding $3$ always divides $S$ (true) and stopping, or believing $43$ must divide $S$ because the sample sequence happens to allow it.
• Miscounting how often each digit appears, e.g. thinking only the hundreds digits matter, which loses the factor $111$.