Difference between revisions of "1995 AJHSME Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | The prime factorization of 6545 is 5 | + | 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 | + | <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 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 contains a three digit number, but we want 6545 to be expressed as ab x cd. 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 |