1996 AJHSME Problems/Problem 1: Difference between revisions

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 $\boxed{B}$ .

1996 AJHSME (Problems • Answer Key • Resources)
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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-b
+==Problem==
+How many positive factors of 36 are also multiples of 4?
+<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6</math>
+==Solution==
+The factors of <math>36</math> are <math>1, 2, 3, 4, 6, 9, 12, 18, </math> and <math>36</math>.
+The multiples of <math>4</math> up to <math>36</math> are <math>4, 8, 12, 16, 20, 24, 28, 32</math> and <math>36</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>.
+==Solution 2==
+<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>.
+==Solution 3==
+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>.
+<math>36 \div 4 = 9</math>, which has <math>3</math> factors, giving us option <math>\boxed{B}</math>.
+== See also ==
+{{AJHSME box|year=1996|before=1995 AJHSME Last Question|num-a=2}}
+* [[AJHSME]]
+* [[AJHSME Problems and Solutions]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

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