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

Latest revision as of 09:12, 1 April 2009

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

@@ Line 1: / Line 1: @@
 ==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
+There exist positive integers <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> with no [[greatest common divisor|common factor]] greater than 1, such that
 <center><math>A\log_{1200} 2+B\log_{1200} 3+C\log_{1200} 5=D.</math></center>
@@ Line 17: / Line 17: @@
 ==See also==
 {{ARML box|year=2006|state=Alabama|num-b=4|num-a=6}}
+[[1995 AHSME Problems/Problem 24|A similar problem]]
-[[1995 AHSME Problems/Problem 24]]
 [[Category:Introductory Algebra Problems]]