1985 AJHSME Problems/Problem 1
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
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 $1\cdot1\cdot1\cdot1\cdot1=\boxed{\textbf{A} 1}.
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.