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

Revision as of 22:16, 9 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, $C=1$ , $B=3$ , $A=2$ . $A+B+C=6 \Rightarrow \mathrm{(A)}$

Revision as of 19:08, 7 January 2008 (view source) Azjps (talk \| contribs) (200 not 2000 ;)) ← Older edit	Revision as of 22:16, 9 January 2008 (view source) Azjps (talk \| contribs) m (1995 AMC 12 Problems/Problem 24 moved to 1995 AHSME Problems/Problem 24) Newer edit →
(No difference)