Best of Math Mole Volume 2

More puzzles, mostly weird ones!

Following the pattern in Best of Math Mole Volume 1, here’s some more problems with comments and solutions (at the end).

2.1: Write down 5 odd numbers that add up to 14.

Just because someone likes math it doesn’t mean he or she doesn’t have other interests.  The problem is one of many where the puzzle writer (not me) showed off some word play to go with the number play. It can be a good problem to use in a set where the problems include heavy and/or tricky calculations.

2.2: A simple calculation. Let [a] be the symbol for a^a so for example [2] = 2^2 = 4. Then let <b> be the symbol for b with b [‘s around it, so that <2> = [[2]] = [4] = 256. Finally, let (c) be the symbol for c with c <‘s around it. Calculate (2).

This is from The Mathematical Experience by Davis and Hersch, which is a great resource for interesting math ideas and a great book to share with the mathematically inclined students. It is hard to classify this problem and I’m not sure the answer is terribly satisfying, but it does open up some interesting conversations about what an answer is, how numbers can be represented, etc.

2.3: tamreF’s Last Theorem: Prove that n^x + n^y = n^z has no solution (x,y,z) in the positive integers for n>2.

This might not be as interesting, but at the time Fermat’s Last Theorem was popular even outside of mathematics and so being able to give students a chance to work with something that at is sort of similar to the FLT was a good thing.

Solutions:

Soln2.1: First you should complain that it is impossible as the sum of 5 odd numbers (by traditional definition) must be odd, but once I stand firm of the claim that there is an answer, you should open up to different ideas for ‘odd.  Then there are various options, like 3, 3, 3, 3, 2, where using 2 as an odd number or being different from 3 is an odd choice.  Let the groans begin.

Soln2.2: We have (2) = <<2>> = <256> = [[...[[256]]...]]] where there should be 256 [‘s in the last expression. That part is easy, but now trying to evaluate it is challenging: [something weird is happening with the formatting]

[256] = 256^{256} \approx 3 \times 10^{316}.

Now raise it to itself! And again and again 252 more times.  I have no idea how big a number this would be or really any way to represent it.  You can try logs, but you really need some kind of repeated log like

\log^n(x) = \log(\log(...(\log(x))...))

with n logs involved. But then we’ve just replaced one precise, but meaningless number with another one (suppose \log^{256} x = 1, what is x?

Soln2.3: First it does hold for n=2 with 2^k + 2^k = 2^{k+1} for any k. So for n>2 what can we do? The clever proof is to look at it in base n and then n^x looks like 10…000 and so the sum of such numbers can only be another such number (in base n) when n=2.  Another approach is to consider w.l.o.g. x \le y < z and then after dividing through by n^x we have 1 + n^s = n^t with 0 \le s < t and then look at the equation mod n and you end up with either 1+0=0 (0<s) or 1+1=0 (0=s) and neither is true for n>2.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s