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*7*11 =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, so every two-digit number pair has to be two number of the form pq. 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 11:42, 21 August 2019
Problem
The number 
can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?
Solution
The prime factorization of is 
, 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:
![\[5*7=35 \text{, } 11*17=187 \text{ X}\]](http://latex.artofproblemsolving.com/d/3/a/d3aa90e7a5d7d66b77a8f8850759521cef81cea1.png)
![\[5*11=55 \text{, } 7*17=119 \text{ X}\]](http://latex.artofproblemsolving.com/a/9/a/a9a993b426bc502c2ff60e6c92e70da519d4a779.png)
![\[5*17=85 \text{, } 7*11=77 \text{ }\surd\]](http://latex.artofproblemsolving.com/8/5/5/855fad6873c017c73a49d5fb538f157cd70f353c.png)
![\[85+77= \boxed{\text{(A)}\ 162}\]](http://latex.artofproblemsolving.com/1/8/f/18f0ea8ccfec284b6d0dff917a81759af971833c.png)
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 | ||
