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 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AHSME Problems and Solutions


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