1984 AHSME Problems/Problem 1

Problem

$\frac{1000^2}{252^2-248^2}$ equals

$\mathrm{(A) \ }62,500 \qquad \mathrm{(B) \ }1,000 \qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ }250 \qquad \mathrm{(E) \ } \frac{1}{2}$

We can use difference of squares to factor the denominator, yielding:

$\frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{1000}$ .

We see that the $1000$ in the denominator cancels with one of the $1000$ s in the numerator, yielding $1000, \boxed{\text{B}}$ .

1984 AHSME (Problems • Answer Key • Resources)
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