Difference between revisions of "2015 AMC 10A Problems/Problem 20"
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Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112. | Also, when adding 4 to 102, you get 106, which has fewer factors than 104, 108, 110, and 112. | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}} | {{AMC10 box|year=2015|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:55, 5 February 2015
Problem
A rectangle has area and perimeter , where and are positive integers. Which of the following numbers cannot equal ?
NOTE: As later announced by the MAA, this problem is incorrect, as it is not stated that the side lengths must be positive integers, so can equal any of the answer choices. This problem was thrown out, and all contestants received full credit for it (whether they answered correctly, incorrectly, or left it blank).
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 fewer factors than 104, 108, 110, and 112.
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.