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

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Thus, <math>\boxed{\text{A}}</math> is the answer.
 
Thus, <math>\boxed{\text{A}}</math> is the answer.
  
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However, if you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}.
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However, if you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}</cmath>.
  
−
That would be </cmath>\frac{10395}{10395}, which is 1. Therefore, the answer is <math>\boxed{\text{A}}</math>
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That would be <cmath>\frac{10395}{10395}</cmath>, which is 1. Therefore, the answer is <math>\boxed{\text{A}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:09, 31 August 2019

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 all of the fractions are equal to one, we have $1\times1\times1\times1\times1$, which is equal to $1$.

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

However, if you want to multiply it out, then it would be \[\frac{15}{99} \times \frac{693}{105}\].

That would be \[\frac{10395}{10395}\], which is 1. Therefore, the answer is $\boxed{\text{A}}$

See Also

1985 AJHSME (Problems • Answer Key • Resources)
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First
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Problem 2
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All AJHSME/AMC 8 Problems and Solutions


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