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

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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:  
 
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>
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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= \boxed{\text{(A)}\ 162}</cmath>
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==See Also==
 +
{{AJHSME box|year=1995|num-b=21|num-a=23}}

Revision as of 02:22, 23 December 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: \[5*7=35 \text{, } 11*17=187 \text{ X}\] \[5*11=55 \text{, } 7*17=119 \text{ X}\] \[5*17=85 \text{, } 7*11=77 \text{ }\surd\] \[85+77= \boxed{\text{(A)}\ 162}\]

See Also

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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