Difference between revisions of "1985 AJHSME Problems/Problem 1"
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==Solution 2 (Brute force)== | ==Solution 2 (Brute force)== | ||
| − | Multlipication gives:<cmath>\frac{15}{99\cdot\frac{693}{105}=\frac{10395}{10395}=\boxed{\text{(A)}\ 1}.</cmath> | + | Multlipication gives:<cmath>\frac{15}{99}\cdot\frac{693}{105}=\frac{10395}{10395}=\boxed{\text{(A)}\ 1}.</cmath> |
==See Also== | ==See Also== | ||
Revision as of 22:45, 12 March 2023
Problem
Solution 1
By the associative property, we can rearrange the numbers in the numerator and the denominator. ![\[\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}\]](http://latex.artofproblemsolving.com/7/4/e/74ebc66bd380cf23e5e3c2a5c8e17afa3e4dd652.png)
Solution 2 (Brute force)
Multlipication gives:![\[\frac{15}{99}\cdot\frac{693}{105}=\frac{10395}{10395}=\boxed{\text{(A)}\ 1}.\]](http://latex.artofproblemsolving.com/7/b/7/7b79b0d5e6124b07c7a9289a3f8b79707dd683ae.png)
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. 
