Difference between revisions of "1985 AJHSME Problems/Problem 1"
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(Note: This method is highly time consuming and should only be used as a last resort in math competitions) | (Note: This method is highly time consuming and should only be used as a last resort in math competitions) | ||
| − | <math> | + | <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 1, or <math>\boxed{\textbf{(A)}\ 1}</math> | ||
| + | |||
| + | ~ lovelearning999 | ||
==Video Solution by BoundlessBrain!== | ==Video Solution by BoundlessBrain!== | ||
Latest revision as of 20:57, 2 October 2024
Contents
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 
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 competitions)
Thus, the answer is 1, or 
~ lovelearning999
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
