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.

(more…)

ShareThis

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!

ShareThis

I’ll be giving away four prizes on Friday - two to the early submitters, one to .mau. for consistently submitting entries to the contest for a really long time, and one to a randomly selected person with a correct submission.

I’ll discuss some of the solutions on Friday but, for now, check out these pictures I made. I came up with this problem by playing with the Fibonacci series and arranging rectangles.

Here is the problem description:

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.

Can you see how these pictures help one to see why the sum is what it is? Do you see why there are two different pictures?

ShareThis

I’ve been playing with this series:

S_n = 1×2² + 2×3² + 3×4² + … + nx(n+1)²

I’ve come up with a formula for determining S_n but I derived the formula in a fairly roundabout way.

I’m interested to know if someone has an elegant way to derive the formula.

ShareThis

If you don’t know how to approach it I suggest looking for a pattern in the sum as you add more terms.

Come on, give it a try! You could win $10.

This time around, since I’ve given you guys a hint, I’ll give three prizes - one each to the two people who have solved it already and one to a randomly other person with a correct solution.

ShareThis

Blinkdagger has announced the winner and the solution for MMM #34. Now, it’s time for MMM #35.

I want to thank Blinkdagger for running 17 contests over the last year and a half. Quan and Daniel have done an excellent job of creating problems and engaging you with fun problem descriptions and great graphics. I’ll miss their participation.

Here’s Monday Math Madness #35:

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.

(more…)

ShareThis

Random.org picked Nikhil Chelliah as the winner for MMM #33. Congratulations, Nikhil!

Here is the problem description:

What’s the prime factorization of the smallest whole number that is divisible by all integers from 1 up to and including 50?

Here is Nikhil’s solution:

The least common multiple of a group of numbers, has, for each prime factor, as many occurrences as the greatest of the group of numbers. Which may make sense iff you already know what I’m talking about.

One algorithm would involve finding the prime factorization of each number up to 50, then taking the most occurrences of each prime factor and producing a final result. A more elegant solution involves identifying all the prime factors without calculating the prime factorization.

The prime factors are simply every prime number up to 50, since all of those numbers are factors, but we don’t care about the composites. Then, for each prime, we ask, what’s the maximum number of times this prime appears in any of the factors?

To answer this we find the greatest exponentiation of the prime number that remains at or below 50. For 2, this is 32; for 3, it’s 27, etc. To find it, we do greatest-integer on the log-base-n of 50.

The following Python code finds all the primes below 50 and their maximum integer exponentiations below 50, and prints each prime that many times.

from math import sqrt, log, floor

def is_prime(n):
if n == 1:
return False
for factor in xrange(2, 1 + sqrt(n)):
if n % factor == 0:
return False
return True

numbers = range(1, 51)
primes = filter(is_prime, numbers)

factors = []
for prime in primes:
for i in xrange( floor( log(50, prime) ) ):
factors.append(prime)
print factors

And the result is:

2, 2, 2, 2, 2, 3, 3, 3, 5, 5, 7, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

(more…)

ShareThis

The solution to MMM #32 is up at Blinkdagger. Here we go with #33.

What’s the prime factorization of the smallest whole number that is divisible by all integers from 1 up to and including 50?

(more…)

ShareThis

Unfortunately our friends Alice and Bob have unknowingly traded their liberty away and are being fenced inside a rectangular pieces of land on the planet Earth, which is assumed to be a perfect sphere of radius 3950 miles.

Alice and Bob are granted one last choice: To decide which plot of land they want to be fenced in.

Both plots of land are bounded by four fences.

Land 1: Two fences run in an exact north-south direction and the other two run in an exact east–west direction. The north-south fences are exactly 10 miles long and east–west fences are exactly 20 miles long.

Land 2: Similar to Land 1, but the north–south fences are 20 miles long and the east–west fences are 10 miles long.

Can you help Alice and Bob to figure out which plot of land has the greater area?

ShareThis

This was a very popular contest. 57 of you sent in your solutions and almost every answer was correct.

Buddha Buck was selected by random.org as the winner of this contest. Here is Buddha’s solution:

Problem: Simplify (1/i)^(1/i) - i^i (where i is sqrt(-1))

Solution:

i*i = -1, -i*i = 1, so 1/i = -i, therefore

(1/i)^(1/i) = (1/i)^(-i)

a^(-b) = (1/a)^b, for all a, b, therefore

(1/i)^(-i) = i^i

So, putting this all together…

(1/i)^(1/i) - i^i
= (1/i)^(-i) - i^i
= i^i - i^i
= 0

(more…)

ShareThis