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

(Solution 1)
(Solution 2 (Brute force))
 
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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>
  
==Soulutions==
+
==Solutions==
 
===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 ([[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.
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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.
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Thus, <math>\boxed{\text{(A)}\ 1}</math> is the answer.
  
==Solution 2 (Brute force)==
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===Solution 2 (Brute force)===
  
If you want to multiply it out, then it would be <cmath>\frac{15}{99} \times \frac{693}{105}</cmath>.
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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)}\ 1}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 12:53, 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, $\boxed{\text{(A)}\ 1}$ 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 $\boxed{\text{(A)}\ 1}$.

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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