1995 AHSME Problems/Problem 24

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 $A + B + C$ ?

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

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

Simplifying and taking the logs away,

$5^A*2^B=200^C=2^{4C}*5^{3C}$ .

Therefore, $A=3C$ and $B=4C$ . Since A, B, and C are relatively prime, C=1, B=4, A=3.

$A+B+C=8 \Rightarrow \mathrm{(C)}$

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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