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

(Solution)
m (Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
The prime factorization of <math>6545</math> is <math>5*7*11*17  =385</math>, 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 <math>6545</math> is <math>5*7*11*17  =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:  
 
<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>
 
<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>
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1995|num-b=21|num-a=23}}
 
{{AJHSME box|year=1995|num-b=21|num-a=23}}

Revision as of 11:54, 26 May 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 $5*7*11*17  =385$, which is a three digit number, but we want 6545 to be expressed as ab x cd . 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 (ProblemsAnswer KeyResources)
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