#4**0 **

Wouldn't it be 1? If you're looking for the **least** positive integer that when factorialed divides 80325, then I think 1 fits the bill.

InterestingUsername Sep 23, 2020

#6**+1 **

I mean, 1! is still 1, and every number can be divided by 1. Also, 1 is positive, and the LEAST positive integer there is. It fits the criteria, but I can agree with you that it seems like a "too easy" answer to the problem. Maybe you could try it?

InterestingUsername Sep 23, 2020

#8**0 **

...wait this is a homework problem...

...homework problems aren't supposed to be put here or answered here.

... I thought this was like a for fun problem on a website or something...

...

InterestingUsername Sep 23, 2020

edited by
InterestingUsername
Sep 23, 2020

edited by InterestingUsername Sep 23, 2020

edited by InterestingUsername Sep 23, 2020

edited by InterestingUsername Sep 23, 2020

edited by InterestingUsername Sep 23, 2020

#9**0 **

no, this is not. It is a practice question. I have done a lot so my score is really high. There higher the harder

DragonXlayer Sep 23, 2020

#10**0 **

Well, ok... the only answer I can think of and prove that fits all the criteria is 1. Sorry I can't be of more help.

InterestingUsername
Sep 23, 2020

#11**0 **

There isn't one. I tried brute force. I divided 80325 by every n! starting at 1! and continuing through 8!. None* of them worked. The only possible answer is 1 and you've already dismissed that, so I don't know what to tell you. Maybe you copied the question incorrectly.

*Actually, they all divided into 80325. The problem doesn't state that the quotient has to be an integer, but in problems like this it is implied.

_{.}

Guest Sep 23, 2020

#15**+1 **

Had you actually tried to solve this question instead of trying to get others to do your work, you would not assume "oh well it is probably a 2 digit number"

For your information, a "two digit number" factorial, the least being 10!, is in the millions.

hugomimihu
Sep 23, 2020

#16**+1 **

An integer can have any number of digits, just so long as it's a whole number, i.e., no fractional component. If that's what you mean. If you're remarking why did I stop at 8! it was because dividing by 8! gives a quotient ~ 1**.**99 so there's no point in dividing by any larger numbers.

_{.}

Guest Sep 23, 2020

#17**0 **

9! is already 360 K or so, no point of going further.

Dragon did not try to solve this problem, not even sure he gave it more than 6 seconds of thought.

I agree with your answer, guest.

hugomimihu
Sep 23, 2020

#13**+1 **

Take prime factorization of this number.

For instance, if you get, 2^49 x 3^39 x 5^10

5! =

2^1 x 3^1 x 2^2 x 5^1, etc etc.

You could use that.

hugomimihu Sep 23, 2020

#14**+1 **

This number's prime factorization is

3^3 × 5^2 × 7 × 17

Anythiing higher than 1! will require a 2, but there is no two, and you're blatantly denying "1",

so...

Other than 1, I see no solution.

hugomimihu Sep 23, 2020

#18

#21**+2 **

1) You said 17 is *incorrect*

3.55687428E14

That is 17!.

I am quite confused as how that is a factor of...

*80325.*

hugomimihu
Sep 23, 2020

#22**0 **

My bad/..... I'm stuck on this too

Find the largest integer n for which 12^n evenly divides 20!.

DragonXlayer Sep 23, 2020

#24**0 **

It appears you would not be able to repeat the previous problem under examination conditions.

Your new question, again, suggests that.

Again:

Prime factorization.

hugomimihu
Sep 24, 2020