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

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==Solution 1==
 
==Solution 1==
By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. <cmath>\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}</cmath>
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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]
  
 
==Solution 2 ==
 
==Solution 2 ==

Revision as of 19:21, 14 December 2023

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

Multlipication gives:\[\frac{15}{99}\cdot\frac{693}{105}=\frac{10395}{10395}=\boxed{\text{(A)}\ 1}.\]

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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