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=?

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- Aug 8th 2010, 05:18 PMgrgrsanjaylogarithms question ♥
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=?

- Aug 8th 2010, 05:28 PMeumyang
$\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.