# Water Box with n Liter

I ran into a basic challenging problem. I see an high school local math Olympiad question.

we have a box that keep n Liter water. each time we extract 1/k Water from box. how many times (minimum) we should do it to see remaining water in box is lower or equal to 1 Liter . (k>2).

I need a creative approach for solving it. :)

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Outline: What about going for non-creative? If we now have an amount $a$ of water, then after one removal we have $a\left(1-\frac{1}{k}\right)$. So after $q$ removals, starting from $n$, we have $$n\left(1-\frac{1}{k}\right)^q.$$ We want $$n\left(1-\frac{1}{k}\right)^q\le 1.$$ Finding the smallest $q$ is a job for the logarithm. I would be more comfortable solving the equivalent inequality $$\left(\frac{k}{k-1}\right)^q\ge n.\tag{1}$$
Added: Use your favourite kind of logarithm. Mine is the natural logarithm (base $e$), but if you really want to, use base $10$. Whatever base $b\gt 1$ we use, call the resulting logarithm by the name $\log$.
Then Inequality (1) can be rewritten as $$q\log\left(\frac{k}{k-1}\right)\ge \log n,$$ or equivalently $$q\ge \frac{\log n}{\log\left(\frac{k}{k-1}\right)}.\tag{2}$$ For brevity, let $w$ be the right-hand side of (2). So we want $q\ge w$. The smallest $q$ that will work is $\lceil w\rceil$, the smallest integer which is $\ge w$.
Use your favourite brand of logarithm. I would suggest natural logarithm (base $e$) but if you really want to use base $10$. We want $q\log(k/(k-1))\ge\log n$, so $q\ge (\log n)/(\log(k/(k-1))$. Call the ugly number we just got by the name $w$. Then actually our answer is $\lceil w\rceil$, the smallest integer which is greater than or equal to $w$. – André Nicolas Aug 31 '14 at 15:43