Difference between revisions of "1985 AJHSME Problems/Problem 1"
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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math> | <math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math> | ||
| − | ==Solution== | + | ==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 ([[Commutative property|commutative property of multiplication]]) so that it looks like <cmath>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}</cmath> | 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 ([[Commutative property|commutative property of multiplication]]) so that it looks like <cmath>\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}</cmath> | ||
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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)== | |
| + | |||
| + | If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}</cmath>. | ||
That would be <cmath>\frac{10395}{10395}</cmath>, which is 1. Therefore, the answer is <math>\boxed{\text{A}}</math> | That would be <cmath>\frac{10395}{10395}</cmath>, which is 1. Therefore, the answer is <math>\boxed{\text{A}}</math> | ||
Revision as of 06:00, 23 May 2020
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 (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.
Solution 2 (Brute force)
If you want to multiply it out, then it would be ![\[\frac{15}{99} \times \frac{693}{105}\]](http://latex.artofproblemsolving.com/a/e/f/aef72f9f5f84337a000af9d5443a857f0592a007.png)
.
That would be ![\[\frac{10395}{10395}\]](http://latex.artofproblemsolving.com/8/c/4/8c47b6f889950cc197fbd97b606ce1ee6c67e2ee.png)
, 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. 
