2007 AMC 12A Problems/Problem 11

The following problem is from both the 2007 AMC 12A #11 and 2007 AMC 10A #22, so both problems redirect to this page.

Problem

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 the 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 factor that always divides $S$ ?

$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$

Solution

A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$ , so $S$ must be divisible by $37\ \mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{(D)}}$ .

Solution 2 (Using the provided sequence)

Since a prime number will always divide $S$ , the prime number must work on any sequence that follows the stated rule. Since the solution kindly provides us with the terms $247,475,756,824$ ,all we have to do is finish the sequence.

One way of finishing the sequence is by adding $562,624$ (before 824) to the sequence and adding them all together. After this, we just divide the result by each of the answer choices. Of the five, $37$ is the largest prime number that divides this sequence, so the answer is ${\boxed{(D)}}$ .

Video Solution

https://www.youtube.com/watch?v=P-phdgVAQqI ~David

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2007 AMC 12A Problems/Problem 11

Contents

Problem

Solution

Solution 2 (Using the provided sequence)

Video Solution

See also