2007 AMC 10A Problem 22

Try a short example first: $123, 231, 312$. Each term starts with the last two digits of the one before, and $312$ leads back to $123$. So this sequence follows the rule.

1. Look at the columns.

The hundreds column holds $1, 2, 3$, and so do the tens and the ones columns, just shifted.
2. Why this always happens. Think of the digits written around a circle. Each term reads three digits in a row. As you move around, each digit is first a ones digit, then a tens digit in the next term, then a hundreds digit in the term after that. Because the list wraps around, this works even at the end.
3. So the sum splits nicely. Let $T$ be the total of the digits around the circle. The hundreds digits add to $T$, the tens digits add to $T$, and the ones digits add to $T$. So
$$S = 100T + 10T + T = 111\,T.$$
4. Factor $111$. $111 = 3 \times 37$. So $37$ always divides $S$.
5. Nothing bigger always works. For $123, 231, 312$ the sum is $666 = 2 \times 3 \times 3 \times 37$. The only choice bigger than $37$ is $43$, and $43$ does not divide $666$.

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

The wrap-around rule means every digit gets used once in each place, so the sum is always $111$ times something. "Always divides" questions have two halves: show the number works for every sequence, then give one example where a bigger number fails. It helps to know $111 = 3 \times 37$ on sight.

Finish the example from the problem into a full loop: $247, 475, 756, 568, 682, 824$. The sum is $3552 = 111 \times 32$. Among the choices, only $3$ and $37$ divide it, and $37$ is bigger. The column argument above then confirms that $37$ works every time.

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 $\boxed{\textbf{(D)}\ 37}$.