Question

Come here : @Jack162

Answer 1

Now proceed from beginning of this concept..

Answer 2

Ok for any prime number p, define $$v_p(n)$$ To be the largest prime that divides n do you see why $$v_p(ab)=v_p(a)+v_p(b)$$

Answer 3

For natural numbers a and b?

Answer 4

Yes I got it..

Answer 5

Ok then we know that $$v_p(m!)=v_p(1)+v_p(2)+v_p(3)+...v_p(m)$$

Answer 6

Yes...

Answer 7

Written more concisely $$v_p(m!)=\sum_{n=1}^m v_p(n)$$

Answer 8

Now suppose p does not divide n

Answer 9

Getting..

Answer 10

What will v_p(n) be?

Answer 11

Don't know..

Answer 12

$$n!=1\times 2\times 3\times 4\times 5\times 6.... \times n$$

Answer 13

How many multiples of p are there in n!?

Answer 14

We have $$p,2p,3p,...p[\frac{n}{p}]$$

Answer 15

Where [.] denotes the floor function

Answer 16

Okay..

Answer 17

do you understand why? the closest multiple of p to n is p[n/p]

Answer 18

Now looking at the rest of the factors in n!, they are not divisible by p so we can exclude them as they have no factor of p.

Answer 19

Thus we need only count the factors of p in$$p*2p*3p*4p*5p*6p*7p*8p*...p[\frac{n}{p}]$$

Answer 20

Okay, getting..

Answer 21

Factoring out p we see that thats equivilent to $$p^{[\frac{n}{p}]}1*2*3*4*5*...*[\frac{n}{p}]$$

Answer 22

Now lets look at the multiples of p in $$1*2*3*4*...*[\frac{n}{p}]$$

Answer 23

OKay..

Answer 24

They are $$p,2p,3p,4p,...p[\frac{n}{p^2}]$$

Answer 25

But sense everything else in the product isn't divisible by p, they each contribute no factors so we can exclude them and only keep the multiples of p.

Answer 26

So again you will take P raise to the power n/p^2 common??

Answer 27

yes, and we then do it with $$[n/p^3],[n/p^4],[n/p^5]$$ So in the end we just need to count how many factors of p, $$p^{[n/p]+[n/p^2]+[n/p^3]...}$$ Has, Which is clearly $$[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+...$$

Answer 28

So we know n! has that many factors of p

Answer 29

Oh, this look similar to formula given by Directrix.

Answer 30

Yes you can use similar reasoning to find the greatest dividing exponent in many other products

Answer 31

So, how to proceed further with this to solve 40!^{40!)

Answer 32

Ok well look if n has say k factors p how many factors will n^a have? it will a times as many factors sense were raising it to the ath power

Answer 33

The same reasoning is used here

Answer 34

Sense were raising 40! to 40!, it will have 40! times the number of factors of 10 40! has

Answer 35

Can you explain it more ??

Answer 36

Ok heres an example, how many factors of 2 does $$(5*2*3*7)^{100}$$ have?

Answer 37

Or how many factors of 5 does, $$(7*5^4*3)^{10}$$ have?

Answer 38

Sense were raising it to a power, it will that times the number of factors on the inside of the expression.

Answer 39

So say we wrote $$40!=10^a*p_1*p_2*p_3$$ How many factors of 10 will $$40!^{40!}$$ have?

Answer 40

Do you see why its 40!a

Answer 41

Assuming none of the primes on the right contribute a factor of 10

Answer 42

Not getting..

Wait, you said the power is the number of factors inside the expression, right??

Answer 43

We want to find the number of trailing zeros in 40!^40!, this is the same thing as counting how many factors of 10 it has

Answer 44

Yes.. This I understand..

Answer 45

Do you see why?

Answer 46

Ok so we need to find out how many factors of 10 it has

Answer 47

Yes,,

Answer 48

Lets say we know how many factors of 10, 40! has call it a, then we can write $$40!=10^{a}*p_1*p_2*p_3*p_4*...$$

Answer 49

Yes we can..

Answer 50

How many factors of 10 does $$40^{40!}=10^{a*40!}*p_1^{40!}*p_2^{40!}...$$ have now?

Answer 51

It has $$40!\times a$$ Factors of 10

Answer 52

Yep..

Answer 53

So we just need to find a, what did we say a was again?

Answer 54

Zeros in 40!

Answer 55

Ok so 40!^{40!} has $$40!\times (\text{# of zeros in 40!)}$$ Factors of 10

Answer 56

I mean factors of 10..

Answer 57

same thing as we said before

Answer 58

Okay, 9 factors of 10 are therer. ie 10^9

Answer 59

So, \(\color{green}{9 \times 40 !}\)

Answer 60

So we know it has $$40!^{40!}$$ has 9*40! trailing zeros

Answer 61

Yes

Answer 62

This is Arithmetic at its best..

Wow..

Can we go with now highest power of 3 in 80! ??

Answer 63

Yes use the formula we proved a while back here, that says how many factors of a prime p n! has

Answer 64

Go back to where we counted how many factors of p n! has

Answer 65

Do you mean

Like this:

\[\frac{80}{3} + \frac{80}{9} + \frac{80}{27} = 26 + 8 + 2 = 36 ??\]

Answer 66

We where each time we take the largest integer no greater then 80/3^j

Answer 67

Yes, thats it

Answer 68

Thanks buddy.. This is quite useful for me..

Answer 69

ostow lawago...

Answer 70

What does it mean @Tease //