Difference between revisions of "1985 AJHSME Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | We '''could''' go at it by just multiplying it out, dividing, etc, but there is a much more | + | We '''could''' go at it by just multiplying it out, dividing, etc, but there is a much more simple method. |
| − | Thus, <math>A</math> is the answer. | + | Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by <math>1</math>, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like <cmath>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}</cmath> |
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| + | Notice that each number is still there, and nothing has been changed - other than the order. | ||
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| + | Finally, since each fraction is equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to <math>1</math>. | ||
| + | |||
| + | Thus, <math>\boxed{\text{A}}</math> is the answer. | ||
Revision as of 17:03, 12 January 2009
Question
Solution
We could go at it by just multiplying it out, dividing, etc, but there is a much more simple method.
Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by 
, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like ![\[\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}\]](http://latex.artofproblemsolving.com/1/9/2/1924081e229a5fced26d3b0d923dea3636de92d5.png)
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 .
Thus, 
is the answer.
