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Difference between revisions of "1985 AJHSME Problems/Problem 1"

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==Solution==
 
==Solution==
  
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We '''could''' go at it by just multiplying it out, dividing, etc, but there is a much more simple method.
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We '''could''' go at it by just [[Multiplication|multiplying]] it out, dividing, etc, but there is a much more simple method.
  
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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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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|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>
  
 
Notice that each number is still there, and nothing has been changed - other than the order.
 
Notice that each number is still there, and nothing has been changed - other than the order.
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==See Also==
 
==See Also==
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[[1985 AJHSME Problems]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 07:12, 6 May 2009

Problem

$\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=$

$\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50$

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 $1$, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like \[\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}\]

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 $1\times1\times1\times1\times1$, which is equal to $1$.

Thus, $\boxed{\text{A}}$ is the answer.

See Also

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