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

(Solution 1)
(Solution 2)
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In our answer choices, the one that is not a factor of <math> 42 </math> is <math> \boxed{\textbf{(D)}\ 28} </math>.
 
In our answer choices, the one that is not a factor of <math> 42 </math> is <math> \boxed{\textbf{(D)}\ 28} </math>.
  
== Solution 2 ==
+
=== Solution 2 ===
 
   
 
   
 
We can look for counterexamples. For example, letting <math>n = 13 \cdot 14 \cdot 15</math>, we see that <math>n</math> is not divisible by 28, so (D) is our answer.
 
We can look for counterexamples. For example, letting <math>n = 13 \cdot 14 \cdot 15</math>, we see that <math>n</math> is not divisible by 28, so (D) is our answer.

Revision as of 18:05, 27 September 2018

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 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 (D) is our answer.

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

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