Difference between revisions of "1996 AJHSME Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
| − | 36 = 4^1 | + | <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 == | ||
Latest revision as of 11:19, 27 June 2023
Problem
How many positive factors of 36 are also multiples of 4?
Solution
The factors of 
are 
and .
The multiples of 
up to are 
and .
Only 
and appear on both lists, so the answer is 
, which is option 
.
Solution 2
. All possible factors of will be here, except for ones divisible by 
and not by . 
. Subtract factors not divisible by , which are 
, 
, and 
. 
, which is .
Solution 3
Divide by , and the remaining factors, when multiplied by , will be factors of .
, which has factors, giving us option .
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
