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

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== Solution ==  
 
== Solution ==  
Factoring the polynomial gives <math>(n+1)(n)(n-1)</math> 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.  
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Factoring the polynomial gives <math>(n+1)(n)(n-1)</math> 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.  Therefore <math>6</math> must divide the given expression.
Multiplying the only factors that can be guaranteed gives <math>3\times2=\boxed{\textbf{(E)} \ 6}</math>
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Plugging in <math>n=2</math> yields <math>6</math>.  So the largest possibility is <math>6</math>.
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Clearly the answer is \boxed{\textbf{(E)} \ 6}$
  
 
== See Also ==
 
== See Also ==

Revision as of 21:49, 9 May 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. Therefore $6$ must divide the given expression. Plugging in $n=2$ yields $6$. So the largest possibility is $6$.

Clearly the answer is \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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