Difference between revisions of "1995 AJHSME Problems/Problem 22"
(→Solution) |
(→Solution) |
||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | The prime factorization of <math>6545</math> is <math>5*7*11*17. 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. 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: |
<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 22:51, 16 October 2018
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:
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 |