Difference between revisions of "1995 AJHSME Problems/Problem 22"

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<math>\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222</math>
 
<math>\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222</math>
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==Solution==
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The prime factorization of 6545 is 5*7*11*17. 5*7*11=385, which is a three digit number, so every two-digit number pair has to be two number of the form pq. Now we do trial and error:
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<cmath>5*7=35 \text{,  } 11*17=187 \text{  X}</cmath> <cmath>5*11=55 \text{,  } 7*17=119 \text{  X}</cmath> <cmath>5*17=85 \text{,  } 7*11=77 \text{ }\surd </cmath> <cmath>85+77=162 \text{(A)}</cmath>

Revision as of 14:32, 5 July 2012

Problem

The number $6545$ can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?

$\text{(A)}\ 162 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 173 \qquad \text{(D)}\ 174 \qquad \text{(E)}\ 222$

Solution

The prime factorization of 6545 is 5*7*11*17. 5*7*11=385, which is a three digit number, so every two-digit number pair has to be two number of the form pq. Now we do trial and error: \[5*7=35 \text{,   } 11*17=187 \text{  X}\] \[5*11=55 \text{,   } 7*17=119 \text{  X}\] \[5*17=85 \text{,   } 7*11=77 \text{ }\surd\] \[85+77=162 \text{(A)}\]