1984 AHSME Problems/Problem 1

Problem

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

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

Solution

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

$\frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{2000}$.

We see that the $1000$ in the denominator cancels with one of the $1000$s in the numerator, yielding $500, \boxed{\text{C}}$.

See Also

1984 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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