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

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==Solution 2==
 
==Solution 2==
Notice that the <math>9 \times 11</math> in the denominator of the first fraction cancels with the same term in the second fraction, and the <math>7</math>s in the numerator and denominator of the second fraction cancel. Then the expression is equal to <math>\frac{3 \times 5}{3 \times 5} = \boxed{\text{(A)} 1}</math>.
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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>.
  
~ cxsmi
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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
  
 
==Video Solution by BoundlessBrain!==
 
==Video Solution by BoundlessBrain!==
 
https://youtu.be/eC_Vu3vogHM
 
https://youtu.be/eC_Vu3vogHM
 
  
 
==See Also==
 
==See Also==

Latest revision as of 20:57, 2 October 2024

Problem

3×59×11×7×9×113×5×7=\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=


(A) 1(B) 0(C) 49(D) 149(E) 50\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50


Solution 1

By the associative property, we can rearrange the numbers in the numerator and the denominator. 335577991111=11111=(A)1\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}

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 (ProblemsAnswer KeyResources)
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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