Difference between revisions of "1995 AHSME Problems/Problem 24"
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==Problems== | ==Problems== | ||
− | There exist positive integers <math>A,B</math> and <math>C</math>, with no common factor greater than 1, such that | + | There exist positive integers <math>A,B</math> and <math>C</math>, with no [[greatest common divisor|common factor]] greater than <math>1</math>, such that |
− | <cmath>A \log_{200} 5 + B \log_{200} 2 = C</cmath> | + | <cmath>A \log_{200} 5 + B \log_{200} 2 = C.</cmath> |
What is <math>A + B + C</math>? | What is <math>A + B + C</math>? | ||
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==Solution== | ==Solution== | ||
− | < | + | <cmath>A \log_{200} 5 + B \log_{200} 2 = C</cmath> |
+ | |||
+ | Simplifying and taking the [[logarithm]]s away, | ||
+ | |||
+ | <cmath>5^A \cdot 2^B=200^C=2^{3C} \cdot 5^{2C}</cmath> | ||
− | + | Therefore, <math>A=2C</math> and <math>B=3C</math>. Since <math>A, B,</math> and <math>C</math> are relatively prime, <math>C=1</math>, <math>B=3</math>, <math>A=2</math>. <math>A+B+C=6 \Rightarrow \mathrm{(A)}</math> | |
− | + | ==See also== | |
+ | {{AHSME box|year=1995|num-b=23|num-a=25}} | ||
− | + | <center>[[2006 Alabama ARML TST Problems/Problem 5|A similar problem]]</center> | |
− | |||
− | + | [[Category:Introductory Algebra Problems]] | |
− | {{ | + | {{MAA Notice}} |
Latest revision as of 13:05, 5 July 2013
Problems
There exist positive integers and , with no common factor greater than , such that
What is ?
Solution
Simplifying and taking the logarithms away,
Therefore, and . Since and are relatively prime, , , .
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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