22 | May | 2008 | Continuous Everywhere but Differentiable Nowhere

Here is my proposed solution to MMM6. (Questions, for the uninitiated or forgetful, are here.)

I actually found these problems to be on the easier side (solved both parts in less than 30 minutes), but I tend to be hasty when I work things out, and overconfident in my thinking, so chances are I’m not right on both counts. So here’s throwing caution to the wind and hoping that I got it right.

(Also, if you liked the problem, don’t forget to look for my generalized problem. Solution to be posted soon.)

Part I Solution:

I created a matrix $X_0=[500 \hspace{12pt} 0]$ which represented the [mayonaise, ketchup] initially. I also created a matrix $Y=[0 \hspace{12pt} 10]$ which represented how much [mayonaise, ketchup] I was adding at each iteration. I let $k=500/510$ which is our “normalization” factor. (You’ll see…)

After the first iteration, we have $X_1=k(X_0+Y)$ which turns out to be $X_1=[490.196 \hspace{12pt} 8.804]$ . You see now what the $k$ does? It makes sure that we have 500 gallons total in the vat.

After the second iteration, we have $X_2=k(k(X_0+Y)+Y)$ which simplifies to $X_2=k^2X_0+(k^2+k)Y$ .

After the third iteration, we have $X_3=k(k(k(X_0+Y)+Y)+Y)$ which simplifies to $X_3=k^3 X_0+(k^3+k^2+k)Y$ .

And you can see the pattern. After the nth iteration, we will have

$X_n=k^n X_0+(k^n+k^{n-1}+...+k^2+k)Y$ .

Now we’re almost done, believe it or not. We can use $k^nX_0=k^n[500 \hspace{12pt} 0]$ to find out how much mayo is in the vat. (Note that since $Y=[0 \hspace{12pt} 10]$ , the $(k^n+k^{n-1}+...+k^2+k)Y$ doesn’t contribute to the mayo.)

We want to find when the mayo is around 250. The mayo after the nth iteration is $500k^n$ . So we have to solve the equation:

$250=500k^n$

A simple manipulation (taking the log of both sides) yields $n\approx 35$ .

Of course a simpler way to think about this is to say that you start out with 500 gallons of mayo. After n iterations, you’ll have $(500/510)^n 500$ gallons of mayo. That brings you to the very last step, which is quickly solvable.

But even though that’s the simpler way to do it, it isn’t the way I started thinking about it. I freely admit that I don’t always find the simplest solution… but it certainly helps when doing generalizations!

What happens if we have add 3 gallons of mayo and 7 gallons of ketchup each time… How many iterations until we get a mixture of 200 gallons of mayo and 300 gallons of ketchup?

My sort-of-involved method works for that! I dare you to try it. You get about 98 iterations. It’s only very slightly tricky, so I’ll write up the solution to that problem in a future post.

Part II Solution

Let’s take the 17 lbs of beef (“it’s what’s for dinner!”).

Let’s have $(single,double)$ represent the number of single and double burgers that Nortrom can eat. We know that these combinations are $(1,8)$ , $(3,7)$ , $(5,6)$ , $(7,5)$ , $(9,4)$ , $(11,3)$ , $(13,2)$ , $(15,1)$ , $(17,0)$ . Let’s see how many different ways that Nortrom can eat one of these combinations of single and double burgers.

Well, let’s look at $(7,5)$ . This is 7 single burgers and 5 double burgers. How many ways are there of ordering them? Clearly there are $\binom{12}{7}$ ways.

Let’s me be clear about this. If Nortrom eats 7 single burgers and 5 double burgers, Nortrom will be eating 12 burgers. We just have to find the number of different ways he can do this. Nortrom, before eating them, placed them down in order of eating them — in 12 slots. Nortrom puts his first burger in slot 1, Nortrom puts his second burger in slot 2, Nortrom puts his third burger in slot 3, etc. Nortrom has to put single burgers in 7 of the 12 slots. (The rest will be filled with double burgers.) The number of ways to do that is $\binom{12}{7}$ . [1]

So then we have, using our combination of single and double burgers listed above: $X=\binom{9}{1}+\binom{10}{3}+\binom{11}{5}+\binom{12}{7}+\binom{13}{9}+\binom{14}{11}+\binom{15}{13}+\binom{16}{15}+\binom{17}{17}=2584$ .

Let’s do exactly the same thing for our 25 lbs of beef!

The combinations are $(1,12)$ , $(3,11)$ , $(5,10)$ , $(7,9)$ , $(9,8)$ , $(11,7)$ , $(13,6)$ , $(15,5)$ , $(17,4)$ , $(19,3)$ , $(21,2)$ , $(23,1)$ , $(25,0)$ .

This leads to a similar conclusion:

$Y=\binom{13}{1}+\binom{14}{3}+\binom{15}{5}+\binom{16}{7}+...+\binom{24}{23}+\binom{25}{25}=121,393$ .

Huzzah!

[1] As a quick aside, note that $\binom{12}{7}=\binom{12}{5}$ . This is because if you have 12 slots and you have to put double burgers in 5 of the slots, then the rest must be filled with single burgers.

Continuous Everywhere but Differentiable Nowhere

I have no idea why I picked this blog name, but there's no turning back now

Day: May 22, 2008

Answer to my generalized MMM6 question

Monday Math Madness 6 Solutions