2007 AMC 12A Problem 11

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.