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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<math>36 = 4^1 \cdot 3^2</math>. All possible factors of <math>36</math> will be here, except for ones divisible by <math>2</math> and not by <math>4</math>. <math>(1+1)\cdot (2+1) = 6</math>. Subtract factors not divisible by <math>4</math>, which are <math>1</math>, <math>3^1</math>, and <math>3^2</math>. <math>6-3=3</math>, which is <math>\boxed{B}</math>.
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==Solution 3==
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Divide <math>36</math> by <math>4</math>, and the remaining factors, when multiplied by <math>4</math>, will be factors of <math>36</math>.
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<math>36 \div 4 = 9</math>, which has <math>3</math> factors, giving us option <math>\boxed{B}</math>.
  
 
== See also ==
 
== See also ==
{{AJHSME box|year=1996|before=First question|num-a=2}}
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{{AJHSME box|year=1996|before=1995 AJHSME Last Question|num-a=2}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 10:19, 27 June 2023

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

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

Solution 3

Divide $36$ by $4$, and the remaining factors, when multiplied by $4$, will be factors of $36$.

$36 \div 4 = 9$, which has $3$ factors, giving us option $\boxed{B}$.

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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