Math Puzzle Help!!!

Akilae729

Alright, I'm helping my girlfriends little brother with a math extra credit problem, and I can't for the life of me figure it out.

The question is as follows

A five digit number is picked from a table of all five digit numbers. What is the probability that the number is evenly divisible by 1,2,3,4,and 5.

I assume that you narrow it down, like all are divisible by one. Even ones are by 2, has to be a 10 or a 5 on the end for a five, cut down the number of even ones by two get get them divisible by four, and then I'm not sure how to handle 3. Is this kinda the right idea?

Any ideas?

Atlus Parker

1/2/3/4/5 = % sounds right assuming no numbers omitted.

Edit: .84% using that solution.

Cauld

I would say 1/1 of the numbers are divisible by 1, 1/2 are divisible by 2, 1/3 are divisible by 3, 1/4 by 4, and 1/5 by five. I believe the answer would then be 1 * 1/2 * 1/3 *1/4 * 1/5 = 1/120 = 0.8%. Someone who's good at math might want to check that though

edit: beated (at least I was right)

SlickShughes

Doesn't feel right to me. All numbers divisible by 4 are also divisible by 2, so you seem to be double counting on that at least. The way I see it, one number in 60 is divisible by all 5 of those, so it's got to be closer to that.

Akilae729

Cauld wrote: »

I would say 1/1 of the numbers are divisible by 1, 1/2 are divisible by 2, 1/3 are divisible by 3, 1/4 by 4, and 1/5 by five. I believe the answer would then be 1 * 1/2 * 1/3 *1/4 * 1/5 = 1/120 = 0.8%. Someone who's good at math might want to check that though

edit: beated (at least I was right)

That makes so much sense, i can't believe i didn't think of that

grungebox

That answer is not correct. It's around 1.67%. Every multiple of 60 qualifies. It's 1500 numbers in that range, total.

EDIT: Confirmed it via MATLAB. Also, you guys are counting numbers with <5 digits, I think.

EDIT 2: The manual way to do it is just find every multiple of 60 in 99999, then subtract the multiples of 60 in 10000. 99999%60-10000%60=1500.

Cauld

no I can fix what I did, I see it already

1/1 are divisible by one,
1/2 are divisble by 2,
1/3 are divisible by 2, but 1/2 of those are divisble by 2
1/4 are divisble by 4, but all of those are divisble by 2 already
1/5 are divisble by 5, but 1/2 of those are divisible by 2, and 1/3 are divisble by 3 (1/2 of these are also divisble by 2)

so you have to take all that into account. This is the part where my math skills fail me

grungebox

Cauld wrote: »

no I can fix what I did, I see it already

1/1 are divisible by one,
1/2 are divisble by 2,
1/3 are divisible by 2, but 1/2 of those are divisble by 2
1/4 are divisble by 4, but all of those are divisble by 2 already
1/5 are divisble by 5, but 1/2 of those are divisible by 2, and 1/3 are divisble by 3 (1/2 of these are also divisble by 2)

so you have to take all that into account. This is the part where my math skills fail me

The exact probability is then 1/60 (1/5*1/4*1/3...no 1/2 since all those are counted in the 1/4 probability). 1/60*89999=# of numbers that are divisible by all of them.

Al_wat

Seems to me it would be 1/3 * 1/4 * 1/5.

1 is contained in all, 2 is contained in 4.

I'm not sure if this is right.

This equals 1 and 2/3rds %, which is the same result Grungebox got.

Atlus Parker

Yep I didn't take into account that 4 is divisible by 4 otherwise my answer would have been 1.66% ie correct.

Gdiguy

grungebox wrote: »

That answer is not correct. It's around 1.67%. Every multiple of 60 qualifies. It's 1500 numbers in that range, total.

EDIT: Confirmed it via MATLAB. Also, you guys are counting numbers with <5 digits, I think.

EDIT 2: The manual way to do it is just find every multiple of 60 in 99999, then subtract the multiples of 60 in 10000. 99999%60-10000%60=1500.

This is the cleanest way to do it - 99999/60 = 1666.65, so there are 1666 multiples of 60 that are less than 99999, and 10000/60=166.6667, so there are 166 multiples of 60 that are less than 10000 (and thus not 5 digits), so you're left with 1500

Demerdar

I believe you guys are double counting BIG time. You need to use the law of Inclusion Exclusion, that should solve the problem.

Gdiguy

Demerdar wrote: »

I believe you guys are double counting BIG time. You need to use the law of Inclusion Exclusion, that should solve the problem.

Who are you talking to? It's pretty easy to confirm that the answer is 1.67% (1500/90000) with a 5 line perl script