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> | ||
| − | == Solution == | + | == Solution 1== |
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. | 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. | ||
Plugging in <math>n=2</math> yields <math>6</math>. So the largest possibility is <math>6</math>. | Plugging in <math>n=2</math> yields <math>6</math>. So the largest possibility is <math>6</math>. | ||
Clearly the answer is <math>\boxed{\textbf{(E)} \ 6}</math> | Clearly the answer is <math>\boxed{\textbf{(E)} \ 6}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | In general, <math>r</math> <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 n. | ||
| + | |||
| + | ~GEOMETRY-WIZARD. | ||
== See Also == | == See Also == | ||
Revision as of 06:25, 31 December 2023
Contents
Problem
The largest number by which the expression 
is divisible for all possible integral values of 
, is:
Solution 1
Factoring the polynomial gives 
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 
must divide the given expression.
Plugging in 
yields . So the largest possibility is .
Clearly the answer is 
Solution 2
In general, 
| 
were and are integers.So here 
| 
- always for any integer n.
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
