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

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<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>
  
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== Solution ==  
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== Solution 1==  
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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.  
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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.
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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 <math>\boxed{\textbf{(E)} \ 6}</math>
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==Solution 2==
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In general, <math>r!</math> | <math>n(n+1)(n+2)...(n+r-1)</math> were <math>r</math> and <math>n</math> are integers. So here <math>3!</math> | <math>n^3</math> - <math>n</math> always for any integer <math>n</math>.Hence,the correct answer is <math>6</math>.
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<math>\boxed{\textbf{(E)} \ 6}</math>
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~geometry wizard.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 06:08, 6 April 2024

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 1

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 $\boxed{\textbf{(E)} \ 6}$

Solution 2

In general, $r!$ | $n(n+1)(n+2)...(n+r-1)$ were $r$ and $n$ are integers. So here $3!$ | $n^3$ - $n$ always for any integer $n$.Hence,the correct answer is $6$. $\boxed{\textbf{(E)} \ 6}$

~geometry wizard.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AHSME Problems and Solutions

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