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: 1500

Small Hint:

Across the whole cycle, each digit serves once as a hundreds digit, once as a tens digit, and once as a units digit

Big Hint:

The sum is $111$ times the total of the units digits, and $111=3\cdot 37$

Solution:

Because of the cycling property, each digit that appears is used the same number of times in the hundreds, tens, and units places.

Let $k$ be the sum of the units digits over all terms. Then $S=100k+10k+k=111k$ $=3\cdot 37\cdot k.$

So $S$ is always divisible by $37.$ It need not be divisible by anything larger: the sequence $123,231,312$ gives $S=666=2\cdot 3^2\cdot 37.$

Thus, the correct answer is D.

2007 AMC 12A Problem 11

Answer: D

Difficulty rating: 1500

Small Hint:

Big Hint:

Solution:

Problem 11 in Other Years