1995 AJHSME Problems/Problem 22: Difference between revisions
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==Solution== | ==Solution== | ||
The prime factorization of <math>6545</math> is <math>5 | 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: | ||
<cmath>5 | <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}} | ||
Revision as of 15:31, 14 July 2021
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, but we want 6545 to be expressed as ab x cd . Now we do trial and error:
![\[5\cdot7=35 \text{, } 11\cdot17=187 \text{ X}\]](http://latex.artofproblemsolving.com/c/5/f/c5f48648c74e82941d47b152c169d0d03ad23af3.png)
![\[5\cdot11=55 \text{, } 7\cdot17=119 \text{ X}\]](http://latex.artofproblemsolving.com/a/6/4/a64bdea4f1a7a518b33456552c44ecbb468e8bab.png)
![\[5\cdot17=85 \text{, } 7\cdot11=77 \text{ }\surd\]](http://latex.artofproblemsolving.com/3/6/1/361ca82bfea52e6778ac6f3687c2611976a7b2bf.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 | ||
