Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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. :)

Math Man Please Help Me !!

share|cite|improve this question
up vote 6 down vote accepted

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 $\ge w$.

share|cite|improve this answer
You are welcome. From the last equation, try to do it on your own. If you tell me what you got, I can tell you whether you are right. – André Nicolas Aug 31 '14 at 15:36
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
dear Nicolas, I'm sorry, but I confused. I don't get how you calculate logarithm. would you please wrote it in a larger space? – Mouna Mokhiab Aug 31 '14 at 16:00
You are right, math in comments is pretty cramped! I have transferred the information to the answer. – André Nicolas Aug 31 '14 at 16:11
thanks so much from your kindly help. – Mouna Mokhiab Aug 31 '14 at 16:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.