Difference between revisions of "1985 AJHSME Problems/Problem 1"
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Thus, <math>\boxed{\text{A}}</math> is the answer. | Thus, <math>\boxed{\text{A}}</math> is the answer. | ||
− | ==Solution 2 (Brute force)== | + | ===Solution 2 (Brute force)=== |
If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}</cmath>. | If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}</cmath>. |
Revision as of 19:46, 24 November 2020
Problem
Solutions
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 (commutative property of multiplication) so that it looks like
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 , which is equal to .
Thus, is the answer.
Solution 2 (Brute force)
If you want to multiply it out, then it would be .
That would be , which is 1. Therefore, the answer is
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.