Math Challenge!

[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)


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.

 

[FoBoT]

42%

 

[emmpee]

0%

EDIT: oops, read it wrong. ignore me.

 

[Semidevil]

I guess 1/2 since I guess there are the same amount of even and odd integers.

 

[amol]

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}

 

[silverpig]

First of all, x must be greater than y to even be considered.

 

[silverpig]

Originally posted by: Amol
haha, 0%? because when you divide a number x by a number y (x: 0<x<1; y: 0<y<1), the number gets larger?

0.8/0.1 = 8

 

[mugs]

I'm going with 25%

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

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.

 

[dionx]

33%

 

[Howard]

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

I'm wondering if he's counting 0 as an integer (and even, at that).

 

[her209]

Assume a, b, c, d:

x = a/b where a < b
y = c/d where c < d

x/y = (a/b)/(c/d) = ad/bc

 

[TuxDave]

read it wrong.. nvm

 

[dionx]

i know 0 is an integer. i just forgot whether its considered even or not

 

[aux]

about 45.467%

 

[silverpig]

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.

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.

That's my point exactly. If x < y, then it's not going to be even.

 

[Zanix]

40%

 

[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)

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

 

[silverpig]

Originally posted by: her209
Assume a, b, c, d:

x = a/b where a < b
y = c/d where c < d

x/y = (a/b)/(c/d) = ad/bc

x and y don't have to be rational though...

 

[FoBoT]

Originally posted by: aux
about 45.467%

so i was close.

sometimes wild guessing is ok

 

[Zanix]

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)

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

Edit: Idiot..

 

[TuxDave]

So we have a consensus that 0 does not count as an even number?

 

[mugs]

Originally posted by: TuxDave
So we have a consensus that 0 does not count as an even number?

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

 

[Goosemaster]

50%

 

[chuckywang]

Originally posted by: aux
about 45.467%

How did you come up with this answer? It's close, but not quite.

 

[TuxDave]

If the solution requires integrals, then I'm on the right track... if not, please stop me.