Difference between revisions of "1994 AJHSME Problems/Problem 10"
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We should list all the positive divisors of <math>36</math> and count them. By trial and error, the divisors of <math>36</math> are found to be <math>1,2,3,4,6,9,12,18,36</math>, for a total of <math>9</math>. However, <math>1</math> and <math>2</math> can't be expressed as <math>N+2</math> for a POSITIVE integer N, so the number of possibilities is <math>\boxed{\text{(A)}\ 7}</math>. | We should list all the positive divisors of <math>36</math> and count them. By trial and error, the divisors of <math>36</math> are found to be <math>1,2,3,4,6,9,12,18,36</math>, for a total of <math>9</math>. However, <math>1</math> and <math>2</math> can't be expressed as <math>N+2</math> for a POSITIVE integer N, so the number of possibilities is <math>\boxed{\text{(A)}\ 7}</math>. | ||
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==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|num-b=9|num-a=11}} | {{AJHSME box|year=1994|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 12:22, 30 October 2016
Problem
For how many positive integer values of 
is the expression 
an integer?
Solution
We should list all the positive divisors of 
and count them. By trial and error, the divisors of are found to be 
, for a total of 
. However, 
and 
can't be expressed as 
for a POSITIVE integer N, so the number of possibilities is 
.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
