Difference between revisions of "2015 AMC 10A Problems/Problem 20"
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So the answer is <math>\boxed{\textbf{(B) }102}</math>. | So the answer is <math>\boxed{\textbf{(B) }102}</math>. | ||
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+ | Also, when adding 4 to 102, you get 106, which has only has 4 factors including 1 and itself whereas the other answer choices have at least 6. (when added to 4). |
Revision as of 12:57, 4 February 2015
Problem
A rectangle has area and perimeter , where and are positive integers. Which of the following numbers cannot equal ?
Solution
Let the rectangle's length and width be and . Its area is and the perimeter is .
Then . Factoring, this is .
Looking at the answer choices, only cannot be written this way, because then either or would be .
So the answer is .
Also, when adding 4 to 102, you get 106, which has only has 4 factors including 1 and itself whereas the other answer choices have at least 6. (when added to 4).