2007 AMC 10A Problem 22

Problem 22 of 25HarderNumber Theory

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?

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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,

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.

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