Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.0 Junior Find the answer

$\frac{1000^2}{252^2-248^2}$ equals
$\mathrm{(A) \ }62,500$ $\mathrm{(B) \ }1,000$ $\mathrm{(C) \ } 500$ $\mathrm{(D) \ }250$ $\mathrm{(E) \ } \frac{1}{2}$

Your answer

Multiple choice: answer with the letter of the option you want.

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 $500, \boxed{\text{C}}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.