Wild About Math! Making Math fun and accessible

20Jul/095

MMM #37: More spiral fun

Our new Monday Math Madness extends the exploration of MMM #36.

Here's the problem:

Based on the introduction to spiral numbers presented in MMM #36, solve one (or both) of these problems:

  1. Come up with an algorithm that tells what number is at an arbitrary X, Y coordinate.
  2. Come up with an algorithm that tells the X, Y coordinates for an arbitrary positive integer.

I'll give three prizes for this contest, one to a random solver as usual, and one each for the best solution to each part of the problem. Even if you've won a prize in the last year you're eligible for one of the "non-random" prizes.

18Jul/091

MMM #36: Spiral Numbers – Winner!


27 of you submitted solutions to MMM #36. Random.org has selected Mathias Malandain as the winner. Congratulations, Mathias!

This was the problem:

Imagine arranging the positive integers in a spiral pattern.
The numbers from 1 to 16 look like this in the spiral pattern.

10  9  8  7
11  2  1  6
12  3  4  5
13 14 15 16

The location of each number corresponds to an X,Y Cartesian coordinate where the number 1 is at the origin: (0,0).
2 is at (-1,0). 3 is at (-1,-1). 4 is at (0,-1). 5 is at (1,-1). 6 is at (1,0). 7 is at (1,1) and so on.

What is the X,Y coordinate of the number 1,000,000?

Show your work.

Most people solved the problem by noticing where squares lie in the spiral.

5Jul/094

MMM #36: Spiral numbers

I'll be contacting the three winners of MMM #35 in the next couple of days to get them their prizes.

Let's move on to MMM #36. I made this one up just for Monday Math Madness!

Imagine arranging the positive integers in a spiral pattern.
The numbers from 1 to 16 look like this in the spiral pattern.

10  9  8  7
11  2  1  6
12  3  4  5
13 14 15 16

The location of each number corresponds to an X,Y Cartesian coordinate where the number 1 is at the origin: (0,0).
2 is at (-1,0). 3 is at (-1,-1). 4 is at (0,-1). 5 is at (1,-1). 6 is at (1,0). 7 is at (1,1) and so on.

What is the X,Y coordinate of the number 1,000,000?

Show your work.

4Jul/091

MMM #35 Fibonacci fun – we have winners!

I'm giving away three prizes this time. Yes, I did say I would give four prizes but when I reviewed the first two submissions one of them was not correct. Henno Brandsma, editor for the Topology Q+A Board, sent the first correct solution, and that was the only correct solution submitted before I gave a small hint. I'm giving a prize to .mau. since he has sent in many solutions to MMMs and has never been selected by random.org. And, I'm giving a prize to Olivier, who was selected by random.org.

Here was the problem:

Let F(0)=1, F(1)=1, and F(n)=F(n-2)+F(n-1). This is the familiar Fibonacci series.
Simplify F(0)xF(1) + F(1)xF(2) + F(2)xF(3) + F(3)xF(4) + … + F(n-1)xF(n) + F(n)xF(n+1)
Show your work.

I got 14 submissions. Most of you sent induction proofs. Since induction doesn't give any insight as to how a solution was derived I can only guess that folks found patterns for sums as n increases then used induction to verify the patterns they saw.

Chao Xu submitted a nice induction proof at the Math Solutions Blog. The solution was password protected to not help others before the submission deadline. I've asked Chao to unprotect it.

Duram had an interesting different-looking but equivalent answer:

S(n+1)=( f(n+2)^2 - f(n) x f(n+1) -1 ) /2

Note how this answer eliminates the need to consider odd and even cases. And, yes, the index is off by 1 in this formula but the insight is excellent.

Jacques Descartes had a clever simplification of the odd/even thing:

S(n)=F(n+1)^2-.5+.5(-1)^n

Do you see how odd/even is handled?

Matthias Malandain proved these two formulas:

S(2n) = (F(2n+1))^2
S(2n-1) = F(2n-1) x F(2n+1)

Another way to approach the problem was by looking at the pictures I posted. While my way does not provide a proof it gives one an intuitive sense of why the answer is what it is.

Watchmath submitted a nice proof that didn't use recursion.

I'm delighted to see the variety in solving this problem.

Stay tuned Monday for the next MMM, right here at Wild About Math!

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