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Math Challenge!

C

chuckywang

Lifer
Two numbers x and y are chosen randomly between 0 and 1. What is the probability that the closest integer to x/y is even?

(From Putnam)


EDIT: 0 is definitely even. Also, please please understand the question before trying to solve it.

FURTHER EDIT: Answer coming tomorrow. For now, here's a hint:

Consider the x-y plane. The square where 0<=x<=1 and 0<=y<=1 is the set of all such points (x,y) that can be chosen. Shade in all points (x,y) such that x/y rounded to the nearest integer is even. Consider the area of this shaded region.

ANSWER: 5/4 - pi/4, Scroll down for solution.
 
haha, i said 0%, but i also forgot that the integers it can be close to are not closed to the set of {0, 1}
 
I'm going with 25%

Originally posted by: silverpig
First of all, x must be greater than y to even be considered.
Click to expand...

Why do you say that?

If x < y, the result is between 0 and 1. So the closest integer would be either 0 or 1. IIRC, 0 is neither even nor odd (I could be wrong), and 1 is most definitely odd. Since in half of all cases x < y, that means you can count 50% out.

Of the remaining 50%, I'm going to assume that half would be closer to an even number and half would be closer to an odd number, therefore 50% * 50% = 25% are closest to an even number.

Edit: My bad, I was thinking of prime numbers. 0 is not prime. It is even. I change my answer to 50%. But that just seems to easy.
 
Originally posted by: mugs
I'm going with 25%

Originally posted by: silverpig
First of all, x must be greater than y to even be considered.
Click to expand...

Why do you say that?

If x < y, the result is between 0 and 1. So the closest integer would be either 0 or 1. IIRC, 0 is neither even nor odd (I could be wrong), and 1 is most definitely odd. Since in half of all cases x < y, that means you can count 50% out.

Of the remaining 50%, I'm going to assume that half would be closer to an even number and half would be closer to an odd number, therefore 50% * 50% = 25% are closest to an even number.
Click to expand...

That's my point exactly. If x < y, then it's not going to be even.
 
Originally posted by: chuckywang
Two numbers x and y are chosen randomly between 0 and 1. What is the probability that the closest integer to x/y is even?

(From Putnam)
Click to expand...

The answer is zero, because there is an infinite number of "numbers" between 0 and 1.
 
Originally posted by: blahblah99
Originally posted by: chuckywang
Two numbers x and y are chosen randomly between 0 and 1. What is the probability that the closest integer to x/y is even?

(From Putnam)
Click to expand...

The answer is zero, because there is an infinite number of "numbers" between 0 and 1.
Click to expand...


Edit: Idiot..
 
Originally posted by: TuxDave
So we have a consensus that 0 does not count as an even number?
Click to expand...

I don't think so, I googled it and came up with a lot of sources saying it was even (because 2*0 = 0)
 
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