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

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==Solution 1==
 
==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 ([[Associative property|associative property of multiplication]]) so that it looks like <cmath>\frac{3}{3} \cdot \frac{5}{5} \cdot \frac{7}{7} \cdot \frac{9}{9} \cdot \frac{11}{11}.</cmath>
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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 [Associative property] so that it looks like <cmath>\frac{3}{3} \cdot \frac{5}{5} \cdot \frac{7}{7} \cdot \frac{9}{9} \cdot \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.

Revision as of 22:37, 12 March 2023

Problem

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

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

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 [Associative property] so that it looks like \[\frac{3}{3} \cdot \frac{5}{5} \cdot \frac{7}{7} \cdot \frac{9}{9} \cdot \frac{11}{11}.\]

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 $1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)}\ 1}$.

Solution 2 (Brute force)

If you want to multiply it out, then it would be \[\frac{15}{99} \cdot \frac{693}{105}= \frac{10395}{10395}= \boxed{\text{(A)}\ 1}.\]

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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