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 13:57, 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).