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Difference between revisions of "2006 Alabama ARML TST Problems/Problem 5"

(New page: ==Problem== There exist positive integers <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> with no common factor greater than 1, such that <center><math>A\log_{1200} 2+B...)
 
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==See also==
 
==See also==
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{{ARML box|year=2006|state=Alabama|num-b=4|num-a=6}}
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[[1995 AHSME Problems/Problem 24]]
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[[Category:Introductory Algebra Problems]]

Revision as of 10:59, 17 April 2008

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

1995 AHSME Problems/Problem 24

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