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Is there easy way of finding power of number in this scenari

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Is there easy way of finding power of number in this scenari

by bfman » Mon Sep 28, 2009 8:06 pm
Hi:

I know. I am sure I read a method somewhere of finding out all the powers of any integer within a certain factorial.

For example, the question:
(30!) / 3^k
What is the greatest integer value for k?

I did it manually and got 14, but I am sure there was a strategy for this. Anyone know?

Thanks!
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there is a theorem

by vivekjaiswal » Mon Sep 28, 2009 10:47 pm
The theorem you are looking for is this

the multiplicity of the prime p occurring in the prime factorization of n! can be expressed exactly as
sum([n/p^i]) for i=1 to infinity, where [] represents the greatest integer function.

https://en.wikipedia.org/wiki/Factorial

HTH,
Vivek
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by bfman » Tue Sep 29, 2009 10:57 am
Thank you Vivek, but I am not too mathematical minded. Can you do a little problem using that formula?

I could be wrong, but I might've come across an easier one. It could be the same though, not sure.

Thanks.
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by ssmiles08 » Tue Sep 29, 2009 4:45 pm
I kind of did it manually too...i don't know if your method is the same as mine.

30! consists of all numbers from 1 to 30. to find the highest power of 3, we need to know how many factors of 3 there are in 30!

I wrote down all factors of 3 till 30 (broken down into primes as well) b/c this question is really only testing for our factors/primes.

3 6 9 12 15 18 21 24 27 30 (you have 10 factors here already)

these numbers are consecutive multiples of 3, so you know that every third term is multiplied by an additional factor of 3, (9, 18, 27)

9 and 18 have one additional 3, 27 has 2 additional 3's. (4 additional 3's)

so it is 10 + 4 = 14 = k
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by gnet » Tue Sep 29, 2009 6:45 pm
(30!) / 3^k

30/3 will give max. 10 multiples
30/(3)^2 will give max. 3 multiples
30/(3)^3 will give max 1. multiple

Add the above to get max. 14 multiples (multiplicity) of prime number 3 in the prime factorization of 30!

---------

Let's try another (a simpler one).

5! / 2^k

5/2 --> max. 2 multiples
5/2^2 --> max. 1 multiple

Thus, prime 2 has a multiplicity of 3 in the prime factorization of 5!

---------

Similarly,

10! / 3^k

10/3 --> max. 3 multiples
10/3^2 --> max. 1 multiple

Thus, prime 3 has a multiplicity of 4 in the prime factorization of 10!
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by vivekjaiswal » Tue Sep 29, 2009 7:21 pm
gnet wrote:(30!) / 3^k

30/3 will give max. 10 multiples
30/(3)^2 will give max. 3 multiples
30/(3)^3 will give max 1. multiple

Add the above to get max. 14 multiples (multiplicity) of prime number 3 in the prime factorization of 30!

---------

Let's try another (a simpler one).

5! / 2^k

5/2 --> max. 2 multiples
5/2^2 --> max. 1 multiple

Thus, prime 2 has a multiplicity of 3 in the prime factorization of 5!

---------

Similarly,

10! / 3^k

10/3 --> max. 3 multiples
10/3^2 --> max. 1 multiple

Thus, prime 3 has a multiplicity of 4 in the prime factorization of 10!
Thanks gnet.
The only thing that is kind of missing in the examples is the greatest integer function. The greatest integer function is defined as follows: [x] is equal to the greatest integer less than or equal to x.
Thus in the above example when we calculate [30/3^2]=[30/9]=[3.33]=3 and similarly for others.

Guys, I hope things are clear with these examples.

Cheers,
Vivek
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by gnet » Sun Oct 25, 2009 7:59 am
Hi, this is an old thread, already well explained by Vivek, but I thought I would throw in here another formula/appproach to such questions, where we need to determine the exact power of a given prime p which divides n! (you will though need to know or learn how to convert a decimal to a different base).

[n - (digit sum of n in base p)] / (p - 1)

For the question above it becomes => (30 - digit sum of 30 in base 3) / (3-1)

30 converted to base 3 is 1010. The digit sum, hence, is 1+0+1+0 = 2.

Therefore, (30 - 2)/(3-1) gives us 28/2 = 14
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