Difference between revisions of "1951 AHSME Problems/Problem 15"

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== Problem ==
 
== Problem ==
The largest number by which the expression <math> n^3 \minus{} n</math> is divisible for all possible integral values of <math> n</math>, is:
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The largest number by which the expression <math> n^3 - n</math> is divisible for all possible integral values of <math> n</math>, is:
  
 
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6</math>
 
<math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6</math>

Revision as of 22:57, 13 March 2015

Problem

The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$, is:

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$

Solution

Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Multiplying the only factors that can be guaranteed gives $3\times2=\boxed{\textbf{(E)} \ 6}$

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AHSME Problems and Solutions

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