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

(New page: ==Problems== There exist positive integers <math>A,B</math> and <math>C</math>, with no common factor greater than 1, such that <cmath>A \log_{200} 5 + B \log_{200} 2 = C</cmath> What is...)
 
(Solution)
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Simplifying and taking the logs away,
 
Simplifying and taking the logs away,
  
<math>5^A*2^B=200^C=2^4C*5^3C</math>.
+
<math>5^A*2^B=200^C=2^{4C}*5^{3C}</math>.
  
 
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=3C</math> and <math>B=4C</math>. Since A, B, and C are relatively prime, C=1, B=4, A=3.

Revision as of 11:32, 7 January 2008

Problems

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 $A + B + C$?


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

Solution

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

Simplifying and taking the logs away,

$5^A*2^B=200^C=2^{4C}*5^{3C}$.

Therefore, $A=3C$ and $B=4C$. Since A, B, and C are relatively prime, C=1, B=4, A=3.

$A+B+C=8 \Rightarrow \mathrm{(C)}$

See Also