Difference between revisions of "1985 AJHSME Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | + | <math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=</math> | |
| − | + | <math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math> | |
| − | ==Solution== | + | ==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] | + | 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 <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>. | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
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| + | == Solution 3 (Brute Force) == | ||
| + | |||
| + | (Note: This method is highly time consuming and should only be used as a last resort in math, also please do not do this.) | ||
| + | <math>3 \times 5 \times 7 \times 9 \times 11 = 10395</math> | ||
| + | |||
| + | <math>9 \times 11 \times 3 \times 5 \times 7 = 10395</math> | ||
| + | |||
| + | Thus, the answer is <math>1</math>, or <math>\boxed{\textbf{(A)}\ 1}</math> | ||
| + | |||
| + | ~ lovelearning999 ~[[shunyipanda]] (Minor edit) | ||
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| + | ==Solution 4 (Easy, similar to solution 1)== | ||
| + | Notice the two fractions are multiplied, so we can consider them as 1 fraction. All terms cancel out, leaving us with <math>\boxed{\textbf{(A)}~1}</math>. | ||
| + | ~[[shunyipanda]] | ||
| + | ==Solution 5 (Capital Pi)== | ||
| + | This fraction can be simplified to <cmath>\frac{\prod_{k=1}^{5}(2k+1)}{\prod_{k=1}^{5}(2k+1)}</cmath> The numerator and denominator are the same, so the answer is <math>\boxed{\textbf{(A)}\ 1}</math>. ~[[shunyipanda]] Note: please can somebody make this look better | ||
| + | |||
| + | ==Video Solution by BoundlessBrain!== | ||
| + | https://youtu.be/eC_Vu3vogHM | ||
==See Also== | ==See Also== | ||
Latest revision as of 13:43, 28 October 2025
Contents
Problem
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 
in the denominator of the first fraction cancels with the same term in the second fraction, the 
s in the numerator and denominator of the second fraction cancel, and the 
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 
.
Solution 3 (Brute Force)
(Note: This method is highly time consuming and should only be used as a last resort in math, also please do not do this.)
Thus, the answer is 
, or 
~ lovelearning999 ~shunyipanda (Minor edit)
Solution 4 (Easy, similar to solution 1)
Notice the two fractions are multiplied, so we can consider them as 1 fraction. All terms cancel out, leaving us with .
~shunyipanda
Solution 5 (Capital Pi)
This fraction can be simplified to ![\[\frac{\prod_{k=1}^{5}(2k+1)}{\prod_{k=1}^{5}(2k+1)}\]](http://latex.artofproblemsolving.com/c/a/6/ca6b4cfb9263656806a1456c6f884423e179d9f2.png)
The numerator and denominator are the same, so the answer is . ~shunyipanda Note: please can somebody make this look better
Video Solution by BoundlessBrain!
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
