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June 2nd, 2015, 10:03 PM   #1
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Factorial

I don't know where to start.
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June 5th, 2015, 07:04 AM   #2
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2 + 3 + 5 + 7 = 17 so n! must be divisible by 2^17, by 3^17, by 5^17 and by 7^17. This implies that you just have to test if it's divisible by 7^17. What is the least n such that n! is divisible by 7^17?
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June 5th, 2015, 11:05 PM   #3
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Quote:
Originally Posted by Hoempa View Post
2 + 3 + 5 + 7 = 17 so n! must be divisible by 2^17, by 3^17, by 5^17 and by 7^17. This implies that you just have to test if it's divisible by 7^17. What is the least n such that n! is divisible by 7^17?
Is there any mathematical/exact way to find the n? I mean if I do it by your way, I have to put the value of n one by one.
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June 5th, 2015, 11:41 PM   #4
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Quote:
Originally Posted by Hoempa View Post
2 + 3 + 5 + 7 = 17 so n! must be divisible by 2^17, by 3^17, by 5^17 and by 7^17. This implies that you just have to test if it's divisible by 7^17. What is the least n such that n! is divisible by 7^17?
EUREKA! I found that the smallest value of n is 105.
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June 6th, 2015, 01:38 AM   #5
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Well done!
You don't have to put in values of n one by one, you can use some method here.

Given a value of n and a prime p, you can compute the multiplicity of p in n!.
Let's do this with your question and answer as an example; find the multiplicity of p = 7 in n! = 105!.

Below (inclusive) 105, (floor(105 / 7) = 15) integers are divisible by 7^1 = 7.
Below (inclusive) 15, (floor(15 / 7) = 2) integers are divisible by 7
(and so 2 integers below (inclusive) 15 * 7 are divisible by 7 * 7 = 7^2 = 49)
Below (inclusive) 2, (floor(2 / 7) = 0) integers are divisible by 7
(and so 0 integers below (inclusive) 2 * 49 are divisible by 49 * 7 = 7^3 = 343)

So the multiplicity of 7 as a primefactor in 105! is 17.

We can use this method to get close to 105 without putting all these values.

Let m = 17 be the multiplicity of n!.
Convert 17 to base 7. This gives
$\displaystyle 23_7$.
So 17 = 2 * 7 + 3.
Now, n is about 17 * 7 = 7(2 * 7 + 3) = 2 * 49 + 3 * 7 (= 119). However, it would have a multiplicity of at least 17. Let's see if we can get a lower value of n.

The one's digit of $\displaystyle 23_7$ is 3 and the 7's digit of $\displaystyle 23_7$ is 2 < 3. So let's subtract the 7's digit from the one's digit. Gives $\displaystyle 21_7 = 15_{10}$. Now, n = 15 * 7 = 105.

This method might need slight modification to always get n exact instead of approximated.
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