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1995 AHSME Problems/Problem 24

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Problems

There exist positive integers and , with no common factor greater than , such that

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

What is ?


\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, and . Since and are relatively prime, , , . 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
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