Difference between revisions of "1995 AHSME Problems/Problem 24"

Revision as of 19:08, 7 January 2008

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

$A \log_{200} 5 + B \log_{200} 2 = C$

What is $A + B + C$ ?

$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$

$A \log_{200} 5 + B \log_{200} 2 = C$

Simplifying and taking the logarithms away,

$5^A \cdot 2^B=200^C=2^{3C} \cdot 5^{2C}$

Therefore, $A=2C$ and $B=3C$ . Since $A, B,$ and $C$ are relatively prime, $C=1$ , $B=3$ , $A=2$ . $A+B+C=6 \Rightarrow \mathrm{(A)}$

@@ Line 1: / Line 1: @@
 ==Problems==
-There exist positive integers <math>A,B</math> and <math>C</math>, with no common factor greater than 1, such that
+There exist positive integers <math>A,B</math> and <math>C</math>, with no [[greatest common divisor|common factor]] greater than <math>1</math>, such that
 <cmath>A \log_{200} 5 + B \log_{200} 2 = C</cmath>
@@ Line 10: / Line 10: @@
 ==Solution==
-<math>A \log_{200} 5 + B \log_{200} 2 = C</math>
+<cmath>A \log_{200} 5 + B \log_{200} 2 = C</cmath>
-Simplifying and taking the logs away,
+Simplifying and taking the [[logarithm]]s away,
-<math>5^A*2^B=200^C=2^{4C}*5^{3C}</math>.
+<cmath>5^A \cdot 2^B=200^C=2^{3C} \cdot 5^{2C}</cmath>
-Therefore, <math>A=3C</math> and <math>B=4C</math>. Since A, B, and C are relatively prime, C=1, B=4, A=3.
+Therefore, <math>A=2C</math> and <math>B=3C</math>. Since <math>A, B,</math> and <math>C</math> are relatively prime, <math>C=1</math>, <math>B=3</math>, <math>A=2</math>. <math>A+B+C=6 \Rightarrow \mathrm{(A)}</math>
-<math>A+B+C=8 \Rightarrow \mathrm{(C)}</math>
+==See also==
+{{Old AMC12 box|year=1995|num-b=23|num-a=25}}
-==See Also==
+[[Category:Introductory Algebra Problems]]
-{{Old AMC12 box|year=1995|num-b=23|num-a=25}}