Question

A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with terms $247, 475,$ and $756$ and end with the term $824.$ Let $S$ be the sum of all the terms in the sequence. What is the largest prime number that always divides $S?$

$3$

$7$

$13$

$37$

$43$

Answer: D

Concepts:place value divisibility

Difficulty rating: 1920

Solution:

Each digit appears as a hundreds digit, a tens digit, and a units digit the same number of times across the sequence.

If $k$ is the sum of the units digits of all terms, then $S = 111k = 3 \cdot 37 \cdot k,$ so $S$ is always divisible by $37.$

The sequence $123, 231, 312$ gives $S = 666 = 2 \cdot 3^2 \cdot 37,$ which has no larger prime factor forced, so $37$ is the answer.

Thus, the correct answer is D.

2007 AMC 10A Problem 22

Answer: D

Difficulty rating: 1920

Solution:

Problem 22 in Other Years