Difference between revisions of "1985 AJHSME Problems/Problem 1"
Rocknroll9 (talk | contribs) m (→Solution) |
Rocknroll9 (talk | contribs) (→Solution) |
||
| Line 13: | Line 13: | ||
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 of the fractions are equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to <math>1</math>. | + | Finally, since all of the fractions are equal to one, we have <math>1\times1\times1\times1\times1</math>, which is equal to <math>1</math>. |
Thus, <math>\boxed{\text{A}}</math> is the answer. | Thus, <math>\boxed{\text{A}}</math> is the answer. | ||
Revision as of 01:07, 24 December 2017
Problem
Solution
We could go at it by just multiplying it out, dividing, etc, but there is a much more simple method.
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 ![\[\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}\]](http://latex.artofproblemsolving.com/1/9/2/1924081e229a5fced26d3b0d923dea3636de92d5.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 
, which is equal to .
Thus, 
is the answer.
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. 
