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2006 Alabama ARML TST Problems/Problem 5

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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

A similar problem

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Alabama_ARML_TST_Problems/Problem_5"

Category: Introductory Algebra Problems

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