Problem:
A rectangle has area A cm2 and perimeter P cm, where A and P are positive integers. Which of the following numbers cannot equal A+P?
Answer Choices:
A. 100
B. 102
C. 104
D. 106
E. 108
Solution:
Let x and y be the lengths of the sides of the rectangle. Then A+P=xy+2x+2y=(x+2)(y+2)−4, so A+P+4 must be the product of two factors, each of which is greater than 2. Because the only factorization of 102+4=106 into two factors greater than 1 is 2⋅53,A+P cannot equal (B)102. Because 100+4=104=4⋅26,104+4=108=3⋅36,106+4=110=5⋅22, and 108+4=112=4⋅28, the other choices equal A+P for rectangles with dimensions 2×24,1×34,3×20, and 2×26, respectively.
The problems on this page are the property of the MAA's American Mathematics Competitions