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== | ||
| + | 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 == | ||
Revision as of 21:47, 5 June 2020
Contents
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
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
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. 
