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

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==Solution==
 
==Solution==
  
The prime factorization of <math>6545</math> is <math>5\cdot7\cdot11\cdot17  =385</math>, which is a three digit number, but we want 6545 to be expressed as ab x cd . Now we do trial and error:  
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The prime factorization of <math>6545</math> is <math>5\cdot7\cdot11\cdot17  =385\cdot17</math>, which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error:  
 
<cmath>5\cdot7=35 \text{,  } 11\cdot17=187 \text{  X}</cmath> <cmath>5\cdot11=55 \text{,  } 7\cdot17=119 \text{  X}</cmath> <cmath>5\cdot17=85 \text{,  } 7\cdot11=77 \text{ }\surd </cmath> <cmath>85+77= \boxed{\text{(A)}\ 162}</cmath>
 
<cmath>5\cdot7=35 \text{,  } 11\cdot17=187 \text{  X}</cmath> <cmath>5\cdot11=55 \text{,  } 7\cdot17=119 \text{  X}</cmath> <cmath>5\cdot17=85 \text{,  } 7\cdot11=77 \text{ }\surd </cmath> <cmath>85+77= \boxed{\text{(A)}\ 162}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1995|num-b=21|num-a=23}}
 
{{AJHSME box|year=1995|num-b=21|num-a=23}}

Latest revision as of 22:31, 11 December 2021

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 $5\cdot7\cdot11\cdot17  =385\cdot17$, which contains a three digit number, but we want 6545 to be expressed as ab x cd. Now we do trial and error: \[5\cdot7=35 \text{,   } 11\cdot17=187 \text{  X}\] \[5\cdot11=55 \text{,   } 7\cdot17=119 \text{  X}\] \[5\cdot17=85 \text{,   } 7\cdot11=77 \text{ }\surd\] \[85+77= \boxed{\text{(A)}\ 162}\]

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AJHSME/AMC 8 Problems and Solutions