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

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bfman: Senior | Next Rank: 100 Posts; Posts: 63; Joined: Tue Mar 31, 2009 3:53 pm

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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Source: Beat The GMAT — Problem Solving | Mon Sep 28, 2009 8:06 pm

vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

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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bfman: Senior | Next Rank: 100 Posts; Posts: 63; Joined: Tue Mar 31, 2009 3:53 pm

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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ssmiles08: Master | Next Rank: 500 Posts; Posts: 472; Joined: Sun Mar 29, 2009 6:54 pm; Thanked: 56 times

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

You got a dream... You gotta protect it. People can't do somethin' themselves, they wanna tell you you can't do it. If you want somethin', go get it. Period.

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gnet: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Thu Apr 16, 2009 8:11 pm; Thanked: 2 times

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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vivekjaiswal: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Sep 15, 2009 7:18 pm; Location: Pune, India; Thanked: 6 times; GMAT Score:630

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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gnet: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Thu Apr 16, 2009 8:11 pm; Thanked: 2 times

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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