2001 AMC 10 Problem 12

Below is the professionally curated solution for Problem 12 of the 2001 AMC 10, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 10 solutions, or check the answer key.

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Concepts:divisibility prime factorization counterexample

Difficulty rating: 1370

12.

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 $n?$

$6$

$14$

$21$

$28$

$42$

Solution:

Among three consecutive integers, at least one is even and one is a multiple of $3,$ so $n$ is divisible by $6.$ With the given factor of $7,$ it is divisible by $6, 14, 21,$ and $42.$

But $28=2^2\cdot7$ requires two factors of $2,$ which is not guaranteed: $5\cdot6\cdot7=210$ is divisible by $7$ but not by $28.$

Thus, the correct answer is D.

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