Difference between revisions of "1994 AJHSME Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | We should list all the positive divisors of 36 and count them. By trial and error, the divisors of 36 are found to be 1,2,3,4,6,9,12,18,36, for a total of 9. However, 1 and 2 can't be expressed as N+2 for POSITIVE integer N, so | + | 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== | ||
| + | {{AJHSME box|year=1994|num-b=9|num-a=11}} | ||
Revision as of 00:29, 23 December 2012
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 | ||
