PRMO 2017 question 5

5. Let $u, v, w$ be real numbers in geometric progression such that $u > v > w$. Suppose
$u^{40} = v^n = w^{60}$. Find the value of $n$.

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If $u,v,w$ form a geometric progression with common ratio $r,$ then we can write

$$v = u r, \quad w = u r^2.$$

The conditions $u^{40} = v^n = w^{60}$ become

$$u^{40} = (ur)^n \quad\text{and}\quad u^{40} = (ur^2)^{60}.$$

From the second equation,

$$(ur^2)^{60} \;=\; u^{60}\,r^{120} \;=\; u^{40},$$

which simplifies (dividing both sides by $u^{40}$) to

$$u^{20}\,r^{120} \;=\; 1 \;\;\Longrightarrow\;\; u^{20} = r^{-120} \;\;\Longrightarrow\;\; u = r^{-6}.$$

Plug this into the first equation,

$$(ur)^n \;=\; (r^{-6}\cdot r)^n \;=\; r^{-6n + n} \;=\; r^{-5n}$$

must equal $u^{40} = (r^{-6})^{40} = r^{-240}.$ Hence

$$r^{-5n}=r^{-240}⟹-5n=-240⟹n=48.$$