Difference between revisions of "1985 AJHSME Problems/Problem 1"
(New page: ==Question== <math>\frac{3\times5}{9\times11} \times \frac{7\times9\times11}{3\times5\times7} </math><br><br> <math>(A) 1 (B) 0 (C) 49 (D) \frac{1}{49} (E) 50</math> ==Solution== We '''co...) |
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==Solution== | ==Solution== | ||
| − | We '''could''' go at it by just multiplying it out, dividing, etc, but there is a much more obvious, simpler method.<br>Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by 1 | + | We '''could''' go at it by just multiplying it out, dividing, etc, but there is a much more obvious, simpler method.<br>Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by 1, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like...<br><br><math>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}</math><br><br>Notice that each number is still there, and nothing has been changed - other than the order.<br>Finally, since each fraction is equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to 1. |
Thus, <math>A</math> is the answer. | Thus, <math>A</math> is the answer. | ||
Revision as of 17:00, 12 January 2009
Question
Solution
We could go at it by just multiplying it out, dividing, etc, but there is a much more obvious, simpler method.
Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by 1, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like...
Notice that each number is still there, and nothing has been changed - other than the order.
Finally, since each fraction is equal to one, we have 
, which is equal to 1.
Thus, 
is the answer.
