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 $A+B+C+D$ .

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

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, $D=1$ , $A=4$ , $B=1$ , $C=2$ . $A+B+C+D=\boxed{8}$