+0

HALP ASAP

0
131
24
+117

What is the least positive integer n such that  n! divides 80325?

Sep 23, 2020

#1
+117
0

Anyoneeee?

Sep 23, 2020
#2
+1

Try 17!

Sep 23, 2020
#3
+117
0

hmmn nope

Sep 23, 2020
#4
+28
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.

Sep 23, 2020
#5
+117
0

No, I don't think that woud work

Sep 23, 2020
#6
+28
+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?

Sep 23, 2020
#7
+117
0

I would, but if I get it wrong it would cost me a lot....

Sep 23, 2020
#8
+28
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...

...

Sep 23, 2020
edited by InterestingUsername  Sep 23, 2020
edited by InterestingUsername  Sep 23, 2020
edited by InterestingUsername  Sep 23, 2020
#9
+117
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

Sep 23, 2020
#10
+28
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.

#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.

.

Sep 23, 2020
#12
+117
0

Well it never said it can't be two digits

DragonXlayer  Sep 23, 2020
#15
+994
+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
+994
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.

hugomimihu  Sep 23, 2020
#13
+994
+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.

Sep 23, 2020
#14
+994
+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.

Sep 23, 2020
#18
+117
0

Well 1 is wrong!

Sep 23, 2020
#19
+994
0

Ok then, tell us the answer.

hugomimihu  Sep 23, 2020
#20
+117
0

its 17....

DragonXlayer  Sep 23, 2020
#21
+994
+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
edited by hugomimihu  Sep 24, 2020
#22
+117
0

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

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

Sep 23, 2020
#23
0

20! mod 12^8 = 0

Guest Sep 23, 2020
#24
+994
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