2006 Alabama ARML TST Problems/Problem 5

Problem

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

$A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.$

Find $A+B+C+D$.

Solution

\[A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D\]

Simplifying and taking the logarithms away,

\[2^A \cdot 3^B \cdot 5^C=1200^D=2^{4D} \cdot 3^{D} \cdot 5^{2D}\]

Therefore, $A=4D$, $B=D$, and $C=2D$. Since $A, B, C,$ and $D$ are relatively prime, $D=1$, $A=4$, $B=1$, $C=2$. $A+B+C+D=\boxed{8}$

See also

2006 Alabama ARML TST (Problems)
Preceded by:
Problem 4
Followed by:
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

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