Difference between revisions of "1984 AHSME Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | We can use difference of squares to factor the denominator, yielding: | + | We can use [[difference of squares]] to factor the [[denominator]], yielding: |
| − | <math> \frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{ | + | <math> \frac{1000^2}{252^2-248^2}=\frac{1000^2}{(252-248)(252+248)}=\frac{1000^2}{(4)(500)}=\frac{1000^2}{2000} </math>. |
| − | We see that the <math> 1000 </math> in the denominator cancels with one of the <math> 1000 </math>s in the numerator, yielding <math> | + | We see that the <math> 1000 </math> in the denominator cancels with one of the <math> 1000 </math>s in the [[numerator]], yielding <math> 500, \boxed{\text{C}} </math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|before=First Problem|num-a=2}} | {{AHSME box|year=1984|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:48, 5 July 2013
Problem
equals
Solution
We can use difference of squares to factor the denominator, yielding:
.
We see that the 
in the denominator cancels with one of the s in the numerator, yielding 
.
See Also
| 1984 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
