logarithms question ♥

May 2010
143
15
chennai,tamil nadu
a,b,c are positive integers forming an increasing G.P and b-a is a perfect cube and \(\displaystyle log_6(a)\)+\(\displaystyle log_6(b)\)+\(\displaystyle log_6(c)\) =6 then a+b+c=?
 
Jan 2010
278
138
\(\displaystyle \begin{aligned}
\log_6\,a + \log_6\,b + \log_6\,c &= 6 \\
6^{\log_6\,a + \log_6\,b + \log_6\,c} &= 6^6 \\
6^{\log_6\,a} \cdot 6^{\log_6\,b} \cdot 6^{\log_6\,c} &= 6^6 \\
abc &= 6^6
\end{aligned}\)

If a, b, and c form a geometric progression then b = ar and c = ar^2 (for some ratio r):
\(\displaystyle \begin{aligned}
abc &= 6^6 \\
a(ar)(ar^2) &= 6^6 \\
a^3 r^3 &= 6^6 \\
ar &= 36 \\
b &= 36
\end{aligned}\)

\(\displaystyle b - a\) has to be a perfect cube, or \(\displaystyle 36 - a\) has to be a perfect cube. There are a few possibilities for a (35, 28, 9), but the only one that will work in our case is if a = 9.

If a = 9 and b = 36, then r = 4, which makes c = 144. So a + b + c = 189.
 
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