I WANT TO DEVIDE THE NUMBER OF 90 INTO 5 PARTS BUT ALL PART SHOULD BE ODD NUMBER.

-- Anonymous, May 08, 2002

Answers

Word.

-- Anonymous, May 08, 2002

Holy shit, did that make me laugh Omar.

-- Anonymous, May 08, 2002

I WANT TO DIVIDE THE GEORGIA LOTTERY INTO EQUAL PARTS AND THEN DEPOSIT IT INTO MY BANK ACCOUNT.

-- Anonymous, May 08, 2002

A 50-cent piece, a quarter and three nickels.

-- Anonymous, May 08, 2002

Sorry Al, you are supposed to DEVIDE those winnings. I'm such a picky bitch.

-- Anonymous, May 08, 2002

Here's the problem: an odd number of odd numbers is always going to add up to be odd. Say, in this case, you have five integers, all odd.

Integer #1 (odd) + Integer #2 (odd) = Sum #1 (even)

Integer #3 (odd) + Integer #4 (odd) = Sum #2 (even)

Integer #5 (odd) + Sum #1 (even) + Sum #2 (even) = Total (odd)

So the only way I can see doing it is if one (or more) of the five is a negative number. But from the formulation of the problem it doesn't sound as if that's an option.

-- Anonymous, May 08, 2002

Wait. No one said there was going to be actual math. No math on MATH!

-- Anonymous, May 08, 2002

Forgive me. I have to go register for the GMAT now.

-- Anonymous, May 08, 2002

Here's the problem: an odd number of odd numbers is always going to add up to be odd.

This holds even in the case of negative numbers.

Proof is simple. A number n is odd if and only if n=2k+1 for some integer k.

Let j and the set of k[1..n] be members of the set of integers. Thus, a sum p of an odd number of odd numbers can be expressed as:

p=sum[i=1,2j+1](2*k[i]+1)

Expanding it out: p=(2k[1]+1) + (2k[2]+1) + (2k[3]+1) + ... + (2k[2j+1] + 1)

Factor: p=2(k[1]+k[2]+...+k[2j+1]) + (2j+1) p=2(k[1]+k[2]+...+k[2j+1]+j) + 1

Let q=k[1]+k[2]+...+k[2j+1]+j. Since k[1..2j+1] and j are all integers, their sum (q) must also be an integer.

Substituting back in, we see that p=2q+1, and is odd by definition.

...

Or you could take the Steven Wright approach: "I was in Vegas and got into a heated debate with the roulette dealer about what I considered to be an odd number."

-- Anonymous, May 08, 2002

Fred, you scare me.

-- Anonymous, May 08, 2002

Fred is like some super-computer-burger-frying MACHINE.

If we could form a Jeopardy team of Fred, WG and the MOC? How much money would they make?

"Alex, we'll take Random Numbers, Obscure Asian Cinema and The History of the American Family for 1,000, please."

-- Anonymous, May 08, 2002

I miss doing proofs. That was fun for me. I do very poorly at Jeopardy and Trivial Pursuit, however.

-- Anonymous, May 09, 2002

Damn you, Frederick, and your Pagan science!!!!

-- Anonymous, May 09, 2002

Word to the F.R.E.D.

-- Anonymous, May 13, 2002