Monday Math Madness 6 Solutions

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.

3 comments

Ξ says:

May 22, 2008 at 7:37 pm

This is neat — I solved the first problem essentially the way you described (the second way) but did the second problem completely differently. What I noticed with the hamburgers was that the numbers formed a Fibonacci Sequence: there was 1 way to eat 1 lb of meat, 2 ways to eat 2 lbs, 3 ways to eat 3 lbs, 5 ways to eat 4 lbs, 8 ways to eat 5 lbs, etc. In general, the number of ways to eat n pounds of meat is the sum of the number of ways to eat (n-1) lbs and the number of ways to eat (n-2) lbs [because you could start by eating a 1lb burger and then eat the remaining (n-1) lbs in however many ways, or you could start by eating a 2-lb burger and then eat the remaining (n-2) lbs in however many ways.].

What I think is really neat is that Pascal’s Triangle ties together our two different solutions to this problem.

The jth entry in the ith row (starting with 0 rather than 1) is i choose j — you can see the beginning in this picture on wikipedia. But if you add diagonals in Pascal’s Triangle, you get Fibonacci numbers, like in this picture. So your formula is just adding those entries of Pascal’s triangle, and my method gave the sum in terms of Fibonacci’s number. I found that referred to as the Fibonacci Binomial Representation on this site (but I’m not sure I can typeset it in the comment — hence the hyperlink).

Thanks for sharing your method — I love it when there are really different ways to look at the same thing.

samjshah says:

May 22, 2008 at 8:53 pm

Oh my gosh, that’s awesome. I too love different solutions to the same problem.

Like this one! Three different solutions, essentially the same, but different ways of approaching it.

https://samjshah.com/2008/05/15/algebraic-manipulation-is-overrated/

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Continuous Everywhere but Differentiable Nowhere

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Monday Math Madness 6 Solutions

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