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1995 AHSME Problems/Problem 24

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Problems

There exist positive integers A,B and C, with no common factor greater than 1, such that

A \log_{200} 5 + B \log_{200} 2 = C.

What is A + B + C?


\mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }

Solution

A \log_{200} 5 + B \log_{200} 2 = C

Simplifying and taking the logarithms away,

5^A \cdot 2^B=200^C=2^{3C} \cdot 5^{2C}

Therefore, A=2C and B=3C. Since A, B, and C are relatively prime, C=1, B=3, A=2. A+B+C=6 \Rightarrow \mathrm{(A)}

See also

1995 AHSME (Problems)
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
A similar problem
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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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