Difference between revisions of "1996 AJHSME Problems/Problem 1"

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Only <math>4, 12</math> and <math>36</math> appear on both lists, so the answer is <math>3</math>, which is option <math>\boxed{B}</math>.
 
Only <math>4, 12</math> and <math>36</math> appear on both lists, so the answer is <math>3</math>, which is option <math>\boxed{B}</math>.
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==Solution 2==
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36 = 4^1 * 3^2. All possible factors of 36 will be here, except for ones divisible by 2 and not by 4. (1+1)*(2+1) = 6. Subtract factors not divisible by 4, which are 1, 3^1, and 3^2. 6-3=3, which is <math>\boxed{B}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 21:47, 5 June 2020

Problem

How many positive factors of 36 are also multiples of 4?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution

The factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18,$ and $36$.

The multiples of $4$ up to $36$ are $4, 8, 12, 16, 20, 24, 28, 32$ and $36$.

Only $4, 12$ and $36$ appear on both lists, so the answer is $3$, which is option $\boxed{B}$.

Solution 2

36 = 4^1 * 3^2. All possible factors of 36 will be here, except for ones divisible by 2 and not by 4. (1+1)*(2+1) = 6. Subtract factors not divisible by 4, which are 1, 3^1, and 3^2. 6-3=3, which is $\boxed{B}$

See also

1996 AJHSME (ProblemsAnswer KeyResources)
Preceded by
1995 AJHSME Last Question
Followed by
Problem 2
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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