Difference between revisions of "2001 AMC 10 Problems/Problem 12"

(Solution)
Line 16: Line 16:
  
 
{{AMC10 box|year=2001|num-b=11|num-a=13}}
 
{{AMC10 box|year=2001|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Revision as of 11:10, 4 July 2013

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 $n$?

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

Solution

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 $\boxed{\textbf{(D)}\ 28}$.

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 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. AMC logo.png