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.

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 10 Problems and Solutions

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2001 AMC 10 Problems/Problem 12

Contents

Problem

Solutions

Solution 1

Solution 2

See Also