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

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==Solution 2==
 
==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>.
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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.
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~geometry wizard.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 05:08, 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
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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