# 2001 AMC 10 Problems/Problem 12

## Problem

Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$. Which of the following is not necessarily a divisor of $n$?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42$

## Solutions

### Solution 1

Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$, meaning it is divisible by $6$.

It also mentions that it is divisible by $7$, so the number is definitely divisible by all the factors of $42$.

In our answer choices, the one that is not a factor of $42$ is $\boxed{\textbf{(D)}\ 28}$.

### Solution 2

We can look for counterexamples. For example, letting $n = 13 \cdot 14 \cdot 15$, we see that $n$ is not divisible by 28, so $\boxed{\textbf{(D) }28}$ is our answer.

### Solution 3(elimination)

No matter what 3 integers you choose, one of them has to be even, so since $14 = 7 \cdot 2$, and it has 7 and 2 as a divisor, answer B is out. Now, if it wasn't divisible by 3, it could be A or C($21 = 7 \cdot 3$,and $6 = 2 \cdot 3$)m so it must be divisible by 3. Therefore, it is either D or E. Since we eliminated 6, if it was E, it would be not divisible by 6($42 = 7 \cdot 6$), but it is not, so the answer is $\boxed{\textbf{(D)}\ 28}$.

~idk12345678

## Video Solution by Daily Dose of Math

~Thesmartgreekmathdude

## See Also

 2001 AMC 10 (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 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.