My Math Forum Factorial

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

June 2nd, 2015, 10:03 PM   #1
Member

Joined: Apr 2015
From: Indonesia

Posts: 53
Thanks: 2

Factorial

I don't know where to start.
Attached Images
 image.jpg (63.5 KB, 32 views)

 June 5th, 2015, 07:04 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 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? Thanks from deadweight
June 5th, 2015, 11:05 PM   #3
Member

Joined: Apr 2015
From: Indonesia

Posts: 53
Thanks: 2

Quote:
 Originally Posted by Hoempa 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.

June 5th, 2015, 11:41 PM   #4
Member

Joined: Apr 2015
From: Indonesia

Posts: 53
Thanks: 2

Quote:
 Originally Posted by Hoempa 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.

 June 6th, 2015, 01:38 AM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 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. Thanks from deadweight

 Tags factorial

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post ducnhuandoan Number Theory 13 June 14th, 2016 12:20 AM bilano99 Calculus 2 May 1st, 2012 10:12 AM panky Algebra 1 November 29th, 2011 12:31 AM thomasthomas Number Theory 13 November 6th, 2010 06:21 AM coolaid317 Calculus 8 April 28th, 2010 06:21 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top