Discrete Mathematics HELP...?

Question

Does anyone know how to solve this .....


Let k= 2 ^ (6) times 3 ^(7)
Find the positive integer n such that k ^ n is a factor of 2006! (2006 factorial), but that k ^(n + 1) is NOT. Fully explain your work.


Help????

Answer 1

Hints:
1. What are all the factors of 2006! ? To know the answer to this, you must know what a factorial is and what a factor is -- I am assuming you know this.
2. What is the smallest number that is NOT a factor of 2006! ?
3. What is the value of k? What value of n will give you a value of k^n which is at least as large as the answer to hint 2?

Answer 2

A difficult question!
Let's find out, first of all, the largest power of 2
dividing 2006! .Then we will do the same for 3.
The largest power of 2 less than 2006 is 1024.
Now we count the multiples of each power of
2 and add them.
There is 1 multiple of 1024
There are 3 multiples of 512
7 multiples of 256
15 multiples of 128
31 multiples of 64
62 multiples of 32
125 multiples of 16
250 multiples of 8
501 multiples of 4
and
1003 multiples of 2.
Total 1998= Highest power of 2 dividing 2006!
Where did the above numbers come from?
Either you can count each one(horrors!)
or note that each one is floor(2006/p^k).
So the largest power of 2^6 dividing 2006! is 1998/6 = 333.
Now do the same for powers of 3.The largest power
of 3 less than 2006 is 729 = 3^6.
If we proceed as above we get the following
for the corresponding numbers:
2 + 8 + 24 + 74 + 222 + 668 = 998.
So the largest power of 3 dividing 2006! is 998.
The largest power of 3^7 dividing 2006! is floor[998/7] = 142.
So the answer to your question is: n = 142.
Hope that helps a bit!