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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...)
 
(→‎Solution 3 (Brute Force))
 
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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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[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
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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.<br>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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[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]
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==Solution 1==
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By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)}  1}[/katex]
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==Solution 2==
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Notice that the <math>9 \times 11</math> in the denominator of the first fraction cancels with the same term in the second fraction, the <math>7</math>s in the numerator and denominator of the second fraction cancel, and the <math>3 \times 5</math> in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with <math>\boxed{\textbf{(A)}~1}</math>.
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~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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== Solution 3 (Brute Force) ==
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(Note: This method is highly time consuming and should only be used as a last resort in math competitions)
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<math>3 \times 5 \times 7 \times 9 \times 11 = 10395</math>
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<math>9 \times 11 \times 3 \times 5 \times 7 = 10395</math>
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Thus, the answer is 1, or <math>\boxed{\textbf{(A)}\ 1}</math>
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~ lovelearning999
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==Video Solution by BoundlessBrain!==
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https://youtu.be/eC_Vu3vogHM
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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}}

Latest revision as of 20:57, 2 October 2024

Problem

[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]


[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex]


Solution 1

By the associative property, we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}[/katex]

Solution 2

Notice that the $9 \times 11$ in the denominator of the first fraction cancels with the same term in the second fraction, the $7$s in the numerator and denominator of the second fraction cancel, and the $3 \times 5$ in the numerator of the first fraction cancels with the same term in the denominator second fraction. Then everything in the expression cancels, leaving us with $\boxed{\textbf{(A)}~1}$.

~cxsmi

Solution 3 (Brute Force)

(Note: This method is highly time consuming and should only be used as a last resort in math competitions)

$3 \times 5 \times 7 \times 9 \times 11 = 10395$

$9 \times 11 \times 3 \times 5 \times 7 = 10395$

Thus, the answer is 1, or $\boxed{\textbf{(A)}\ 1}$

~ lovelearning999

Video Solution by BoundlessBrain!

https://youtu.be/eC_Vu3vogHM

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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