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1989 AHSME Problems/Problem 17

Problem

The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989 \ \text{cm}$. The length of each side of the triangle exceeds the length of each side of the square by $d \ \text{cm}$. The square has perimeter greater than 0. How many positive integers are NOT possible value for $d$?

$\text{(A)} \ 0 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 221 \qquad \text{(D)} \ 663 \qquad \text{(E)} \ \infty$

Solution

If t is the side of the triangle, and s is the side of the square, then

\[3t-4s=1989\] \[t-s=d\]

Solving the first equation for t gives

\[t = \frac{4s+1989}{3}\]

Substituting into the second equation,

\[\frac{4s+1989}{3} - s = d\] \[\frac{s+1989}{3} = d\] \[s+1989 = 3d\]

If s = 0, d = 663. But s has to be greater than 0, so the first 663 positive integers aren't possible $\to\boxed{\textbf{(D)}}$

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AHSME Problems and Solutions

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