Difference between revisions of "1985 AJHSME Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
− | Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by <math>1</math>, we can rearrange the numbers in the numerator and the denominator | + | Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by <math>1</math>, we can rearrange the numbers in the numerator and the denominator [Associative property] so that it looks like <cmath>\frac{3}{3} \cdot \frac{5}{5} \cdot \frac{7}{7} \cdot \frac{9}{9} \cdot \frac{11}{11}.</cmath> |
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. |
Revision as of 22:37, 12 March 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] 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 .
Solution 2 (Brute force)
If you want to multiply it out, then it would be
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.