Difference between revisions of "1994 AJHSME Problems/Problem 10"
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<math>\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12</math> | <math>\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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 equal to <math>N+2</math> for a POSITIVE integer N, so the number of possibilities is <math>\boxed{\text{(A)}\ 7}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | To find the number of positive divisors of <math>36</math>, first prime factorize <math>36</math> to get <math>2^2*3^2</math>. Then add <math>1</math> to the power of both <math>2</math> and <math>3</math> to get <math>3</math>. Multiply <math>3*3</math> to get <math>9</math>. Since the problem is asking only for positive integer values of N, subtract <math>2</math> from <math>9</math> (since <math>2-2</math> and <math>1-2</math> result in integers that are not positive) to get <math>\boxed{\text{(A)}\ 7}</math>. | ||
+ | |||
+ | ~ spoamath321 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1994|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 07:42, 26 May 2021
Contents
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 equal to for a POSITIVE integer N, so the number of possibilities is .
Solution 2
To find the number of positive divisors of , first prime factorize to get . Then add to the power of both and to get . Multiply to get . Since the problem is asking only for positive integer values of N, subtract from (since and result in integers that are not positive) to get .
~ spoamath321
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.