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

(Solution 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 ([[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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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} \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 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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Finally, since all of the fractions are equal to one, we have $1\cdot1\cdot1\cdot1\cdot1=\boxed{\textbf{A} 1}.
 
 
Thus, <math>\boxed{\text{(A)}\ 1}</math> is the answer.
 
  
 
==Solution 2 (Brute force)==
 
==Solution 2 (Brute force)==

Revision as of 19:59, 1 February 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 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\cdot1\cdot1\cdot1\cdot1=\boxed{\textbf{A} 1}.

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

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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