Wild About Math! Making Math fun and accessible


Now that the MMM #35 deadline has passed …

MMM #35 turned out to be harder than I thought judging by the small number of submissions I received.

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?

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Can you sum this?


I've been playing with this series:

Sn = 1x22 + 2x32 + 3x42 + ... + nx(n+1)2

I've come up with a formula for determining Sn 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.

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MMM #35: Is it too hard?

I've only gotten two solutions to the "Fibonacci Fun" problem I proposed. Is everyone on vacation and not thinking about Math, do people find the problem boring or is it too hard? I don't think it's one of my harder problems. I want to state that I approached the problem in a different way than each of the people who provided solutions. So, there are at least three ways to approach the problem.

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.

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MMM #35: Fibonacci Fun

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.


MMM #33: We have a winner!

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) ) ):
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