Difference between revisions of "1985 AJHSME Problem 1"

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==Problem==
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This page is a duplicate of this [https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_1 webpage].
 
 
<math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=</math>
 
 
 
<cmath>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)} \frac{1}{49}\ \qquad \text{(E)}\ 50</cmath>
 
 
 
==Solution==
 
 
 
<math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}= \frac{3 \times 5 \times 7 \times 9 \times 11}{9 \times 11 \times 3 \times 5 \times 7} = \boxed{1}</math>
 
 
 
The answer is <math>\text{(A) 1}.</math>
 
 
 
 
 
 
 
==Solution==
 
 
 
the numeretor is 3*5*7*9*11, so is the denominator so (3*5*7*9*11)/(3*5*7*9*11)=1 
 
 
 
 
 
-mathmax12
 
 
 
==Video Solution==
 
https://youtu.be/dszCk0HVWH8
 
 
 
~savannahsolver
 
 
 
== See Also ==
 
{{AJHSME box|year=1985|num-b=0|num-a=2}}
 
{{MAA Notice}}
 

Latest revision as of 21:24, 12 March 2023

This page is a duplicate of this webpage.