Difference between revisions of "1985 AJHSME Problems/Problem 1"

(Solution 2 (Brute force))
(Solution 1)
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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>.
 
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{\mathbf{(A )}1}</math> is the answer.
  
 
===Solution 2 (Brute force)===
 
===Solution 2 (Brute force)===

Revision as of 13:51, 22 December 2020

Problem

$\frac{3\times 5}{9\times 11}\times \frac{7\times 9\times 11}{3\times 5\times 7}=$

$\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50$

Solutions

Solution 1

Noticing that multiplying and dividing by the same number is the equivalent of multiplying (or dividing) by $1$, we can rearrange the numbers in the numerator and the denominator (commutative property of multiplication) so that it looks like \[\frac{3}{3} \times \frac{5}{5} \times \frac{7}{7} \times \frac{9}{9} \times \frac{11}{11}.\]

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 $1\times1\times1\times1\times1$, which is equal to $1$.

Thus, $\boxed{\mathbf{(A )}1}$ is the answer.

Solution 2 (Brute force)

If you want to multiply it out, then it would be \[\frac{15}{99} \times \frac{693}{105}.\]

That would be \[\frac{10395}{10395},\] which is 1. Therefore, the answer is $\boxed{\mathbf{(A )}1}$

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
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


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