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avatar+118 

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

 Sep 23, 2020
 #1
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0

Anyoneeee?

 Sep 23, 2020
 #2
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+1

Try 17!

 Sep 23, 2020
 #3
avatar+118 
0

hmmn nope

 Sep 23, 2020
 #4
avatar+28 
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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
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No, I don't think that woud work

 Sep 23, 2020
 #6
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+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
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I would, but if I get it wrong it would cost me a lot....

 Sep 23, 2020
 #8
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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
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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
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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
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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
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0

Well it never said it can't be two digits

DragonXlayer  Sep 23, 2020
 #15
avatar+986 
+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
avatar
+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
avatar+986 
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
avatar+986 
+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
avatar+986 
+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
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0

Well 1 is wrong!

 Sep 23, 2020
 #19
avatar+986 
0

Ok then, tell us the answer.

hugomimihu  Sep 23, 2020
 #20
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0

its 17....

DragonXlayer  Sep 23, 2020
 #21
avatar+986 
+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
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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
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0

20! mod 12^8 = 0

Guest Sep 23, 2020
 #24
avatar+986 
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

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