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

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==Problem==
 
==Problem==
  
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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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[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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<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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==Solutions==
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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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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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==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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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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==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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Thus, <math>\boxed{\text{A}}</math> is the answer.
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~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
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===Solution 2 (Brute force)===
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== Solution 3 (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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(Note: This method is highly time consuming and should only be used as a last resort in math competitions)
  
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That would be <cmath>\frac{10395}{10395}</cmath>, which is 1. Therefore, the answer is <math>\boxed{\text{A}}</math>
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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
  
 
==See Also==
 
==See Also==

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)
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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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