Difference between revisions of "1985 AJHSME Problem 1"
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==Problem== | ==Problem== | ||
| − | <math>\frac{ | + | <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> | <cmath>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)} \frac{1}{49}\ \qquad \text{(E)}\ 50</cmath> | ||
Revision as of 13:58, 19 February 2023
Problem
![\[\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)} \frac{1}{49}\ \qquad \text{(E)}\ 50\]](http://latex.artofproblemsolving.com/e/0/b/e0b568a124650ef62e9b036108fb434b7d607133.png)
Solution
The answer is 
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
~savannahsolver
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 0 |
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. 
