Difference between revisions of "1985 AJHSME Problems/Problem 1"
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| − | == | + | ==Problem== |
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| − | = | + | <math>\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=</math> |
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| − | 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 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> | + | <math>\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50</math> |
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| + | ==Solutions== | ||
| + | ===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> | ||
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 | + | 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>. |
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| + | Thus, <math>\boxed{\text{(A)}\ 1}</math> is the answer. | ||
| + | |||
| + | ===Solution 2 (Brute force)=== | ||
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| + | If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}.</cmath> | ||
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| + | That would be <cmath>\frac{10395}{10395},</cmath> which is 1. Therefore, the answer is <math>\boxed{\text{(A)}\ 1}</math>. | ||
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| + | ==See Also== | ||
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| + | {{AJHSME box|year=1985|before=First <br> Question|num-a=2}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
| + | |||
| − | + | {{MAA Notice}} | |
Revision as of 13:53, 22 December 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 ![\[\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}.\]](http://latex.artofproblemsolving.com/1/c/0/1c02d21c6530ff19d1539b9ef03316a676c8a369.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/e/a/0/ea0cbfa56a4ff8ce9ddca6a6127eff51b700e0d8.png)
That would be ![\[\frac{10395}{10395},\]](http://latex.artofproblemsolving.com/8/f/a/8fae9d464ee2f67258053ad8f326a2ecd34d8121.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. 
