+117 What is the least positive integer n such that n! divides 80325?
+28 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.
+28 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?
+28 ...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...
...
+117 no, this is not. It is a practice question. I have done a lot so my score is really high. There higher the harder
+28 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.
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.
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+1008 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.
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.
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+1008 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.
+1008 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.
+1008 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.
+1008 1) You said 17 is incorrect
3.55687428E14
That is 17!.
I am quite confused as how that is a factor of...
80325.
+117 My bad/..... I'm stuck on this too
Find the largest integer n for which 12^n evenly divides 20!.
+1008 It appears you would not be able to repeat the previous problem under examination conditions.
Your new question, again, suggests that.
Again:
Prime factorization.
