Difference between revisions of "1985 AJHSME Problems/Problem 1"
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Notice that each number is still there, and nothing has been changed - other than the order. | Notice that each number is still there, and nothing has been changed - other than the order. | ||
| − | Finally, since all of the fractions are equal to one, we have <math>1\cdot1\cdot1\cdot1\cdot1=\boxed | + | Finally, since all of the fractions are equal to one, we have <math>1\cdot1\cdot1\cdot1\cdot1=\boxed{(A)~1}</math>. |
==Solution 2 (Brute force)== | ==Solution 2 (Brute force)== | ||
Revision as of 20:15, 1 February 2023
Problem
Solution 1
Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by 
, we can rearrange the numbers in the numerator and the denominator (associative property of multiplication) so that it looks like ![\[\frac{3}{3} \cdot \frac{5}{5} \cdot \frac{7}{7} \cdot \frac{9}{9} \cdot \frac{11}{11}.\]](http://latex.artofproblemsolving.com/2/7/e/27e8fbc2ea1c2427988b2b248bc1d64f03efcf94.png)
Notice that each number is still there, and nothing has been changed - other than the order.
Finally, since all of the fractions are equal to one, we have 
.
Solution 2 (Brute force)
If you want to multiply it out, then it would be ![\[\frac{15}{99} \cdot \frac{693}{105}= \frac{10395}{10395}= \boxed{\text{(A)}\ 1}.\]](http://latex.artofproblemsolving.com/8/8/0/88084cf50a5490c23654f40c006557892d2cc3b7.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. 
