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Difference between revisions of "1994 AJHSME Problems/Problem 10"

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==Solution==
 
==Solution==
  
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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 there are 7 possibilities. <math>\text{(A)}</math>
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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 equal to <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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==Solution 2==
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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>.
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~ spoamath321
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==See Also==
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{{AJHSME box|year=1994|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 07:42, 26 May 2021

Problem

For how many positive integer values of $N$ is the expression $\dfrac{36}{N+2}$ an integer?

$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

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 equal to $N+2$ for a POSITIVE integer N, so the number of possibilities is $\boxed{\text{(A)}\ 7}$.

Solution 2

To find the number of positive divisors of $36$, first prime factorize $36$ to get $2^2*3^2$. Then add $1$ to the power of both $2$ and $3$ to get $3$. Multiply $3*3$ to get $9$. Since the problem is asking only for positive integer values of N, subtract $2$ from $9$ (since $2-2$ and $1-2$ result in integers that are not positive) to get $\boxed{\text{(A)}\ 7}$.

~ 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

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