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

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==Question==
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==Problem==
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<math>\frac{3\times5}{9\times11} \times \frac{7\times9\times11}{3\times5\times7} </math><br><br>
 
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<math>(A) 1 (B) 0 (C) 49 (D) \frac{1}{49} (E) 50</math>
 
  
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==Solution==
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<math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=</math>
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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.
 
  
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Thus, <math>A</math> is the answer.
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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math>
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==Solution 1==
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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>
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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 all of the fractions are equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to <math>1</math>.
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Thus, <math>\boxed{\text{(A)}\ 1}</math> is the answer.
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==Solution 2 (Brute force)==
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If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}.</cmath>
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That would be <cmath>\frac{10395}{10395},</cmath> which is 1. Therefore, the answer is <math>\boxed{\text{(A)}\ 1}</math>.
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==See Also==
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{{AJHSME box|year=1985|before=First <br> Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 02:13, 16 February 2021

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 1

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, $\boxed{\text{(A)}\ 1}$ is the answer.

Solution 2 (Brute force)

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 $\boxed{\text{(A)}\ 1}$.

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
First
Question 		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
All AJHSME/AMC 8 Problems and Solutions


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