Difference between revisions of "2015 AMC 10A Problems/Problem 20"
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I am afraid the problem has an error: the actual sides of the rectangle had to be integers. As stated, every answer choice would work, with one of the sides being <math>2</math>, and the other, a half-integer. E.g., for <math>102</math>, the sides of the rectangle would be <math>2</math> and <math>49/2</math>. | I am afraid the problem has an error: the actual sides of the rectangle had to be integers. As stated, every answer choice would work, with one of the sides being <math>2</math>, and the other, a half-integer. E.g., for <math>102</math>, the sides of the rectangle would be <math>2</math> and <math>49/2</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}} | ||
+ | {{MAA Notice}} |
Revision as of 22:16, 4 February 2015
Contents
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 less factors than 104, 108, 110, and 112.
No Solution
I am afraid the problem has an error: the actual sides of the rectangle had to be integers. As stated, every answer choice would work, with one of the sides being , and the other, a half-integer. E.g., for
, the sides of the rectangle would be
and
.
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.