Auto-Tune (up)
Logic; Difficulty: easy
The Riddler: The Auto Tune up
I’m late to this week’s Riddler Classic but figured I’d post this anyway since the limiting case might be interesting.
You want to change the transmission fluid in your old van, which holds 12 quarts of fluid. At the moment, all 12 quarts are “old.” But changing all 12 quarts at once carries a risk of transmission failure.
Instead, you decide to replace the fluid a little bit at a time. Each month, you remove one quart of old fluid, add one quart of fresh fluid and then drive the van to thoroughly mix up the fluid. (I have no idea if this is mechanically sound, but I’ll take Travis’s word on this!) Unfortunately, after precisely one year of use, what was once fresh transmission fluid officially turns “old.”
You keep up this process for many, many years. One day, immediately after replacing a quart of fluid, you decide to check your transmission. What percent of the fluid is old?
Added variations I consider:
- What is the limiting case if you were to do this n times a year as n becomes really ‘large’?
- What about if you start with 12 fresh quarts and replace 1 quart each month as before? What percent would be old after a sufficiently long time?
*spoiler alert* solution and answer follow below
Getting right into it
This is a relatively easy problem. \( \displaystyle \frac{1}{12} \) of any part of the existing fluid is depleted each month. So if we start with all Old fluid, after the first replacement in month 1 you have \(\displaystyle \left(\frac{11}{12}\right )\) of the Old fluid remaining and \(\displaystyle \left(\frac{1}{12}\right )\) New. With the next change in month 2 , you have \(\displaystyle \left(\frac{11}{12}\right )^2\) of the Old, and \(\displaystyle \left(\frac{11}{12}\right) \left(\frac{1}{12}\right) \) of New but aged level 1 fluid and \(\displaystyle \left(\frac{1}{12}\right )\) New.
Continuing in this vein, you have after 12 months, only \(\displaystyle \left(\frac{11}{12}\right )^{12} \) of the Old fluid is remaining and everything else is New, albeit in varying stages of aging. The closest to being Old (out of the total 12) is the very first quart new that was added. This has also been depleted at the same rate of \( \displaystyle \frac{1}{12} \) each month/change. So you are left with \(\displaystyle \left(\frac{11}{12}\right )^{11}.\left(\frac{1}{12}\right) \) almost Old. There’s \(\displaystyle \left(\frac{11}{12}\right )^{10}\left(\frac{1}{12}\right) \) of the next level and so on…with \(\displaystyle \left(\frac{1}{12}\right )\) New.
With the next change you have \(\displaystyle \left(\frac{11}{12}\right )^{12} \) of the original Old and \(\displaystyle \left(\frac{11}{12}\right )^{12}.\left(\frac{1}{12}\right) \) the newly minted Old from the almost Old for a total Old proportion of
\begin{align*}
\displaystyle \left(\frac{11}{12}\right )^{13} + \displaystyle \left(\frac{11}{12}\right )^{12}.\left(\frac{1}{12}\right) &= \left(\frac{11}{12}\right )^{12}\left [\frac{11}{12} + \frac{1}{12} \right ] \\
&= \left(\frac{11}{12}\right )^{12}
\end{align*}
Each of the other varying stages have moved to one higher level of aging. There’s \(\displaystyle \left(\frac{11}{12}\right )^{10}.\left(\frac{1}{12}\right) \) of the next level and so on…with \(\displaystyle \left(\frac{1}{12}\right )\) New same as the step before implying steady state has been reached.
So the answer here is \(\displaystyle \left(\frac{11}{12}\right )^{12} = 35.2\%\) from here on out!
Generalizing:
It follows by extension of the above reasoning that if we replace \( \displaystyle \frac{1}{n} \) of the fluid each time, the fraction of Old fluid remaining after change \(n + 1\) is steady at\(\displaystyle \left(\frac{n-1}{n}\right )^{n}\).
1. The limiting case
If you decided to change \( \displaystyle \frac{1}{365} \) of the fluid each morning, after a year the fraction of Old left is \(36.74\%\).
If we take this to the extreme:
\begin{align*}
\textit{As } n &\rightarrow \infty \\ \\
\displaystyle \left(\frac{n-1}{n}\right )^{n} &\rightarrow \displaystyle \left(1 – \frac{1}{n}\right )^{n} \\ \\
&= \frac{1}{e} \\ \\
&=36.79\%
\end{align*}
2. Starting with all New
It’s fairly easy to show using the same reasoning as with the starting with the Old case, that if you start with all New fluid, you still end up with the exact same proportion of Old after a year of changes!