Question

Could anyone explain why in 1^1 * 2^2 * 3^3 * ... * 25^25 valuation of each prime can be rewritten as: (latex in the comments)

Answer 1

\(n = 1^1 \cdot 2^2 \cdot 3^3 \cdots 25^{25} \) value of each prime can be rewritten as:
\(v_p(n)=v_p(\prod_{k = 1}^{25}{k^k}) = \sum_{k = 1}^{25}{v_p(k^k)} = \sum_{k = 1}^{25}{v_p(k) \cdot k}\)

I can't understand how this has been built.

Answer 2

Source of the solution & problem: http://math.stackexchange.com/questions/1839495/find-last-three-non-zero-digits-of-11-cdot-22-cdot-33-cdot-cdot-25/1839655#1839655

Answer 3

@ParthKohli

Answer 4

What do you mean by `valuation` ?

Answer 5

It's probably the power of each prime in the factorisation.

So\[v_p \left(\prod_{1\le k \le 25} k^k\right) = \sum_{k=1}^{25} v_p(k^k) \]Literally translates to the sum of powers in the product is the sum of powers in each term, etc.

Answer 6

Although I can't be too sure.
Let me look at the link

Answer 7

Okay, nice. So he is writing the prime factorization of the given number first.

Answer 8

(By calculating the exponent of each prime in the given number)

Answer 9

Well, I understand what this does (in a way), but I can't seem to understand how it works.
Why after the second equal sign there is \(v_p(k^k)\)?

Answer 10

Zyberg maybe first forget what he is doing and try to figure out the prime factorization on your own.

After that go through his method and you will understand. He is just using fancy sum and product symbols.

Answer 11

It's the sum of all \(v_p (k^k)\)s from 1 to 25.

So for example, \(v_2 (5\cdot 6\cdot 7) =v_2 (210) = 1\).
Also \(v_2 (5) + v_2 ( 6) + c_2(7) =0+1+0 = 1\).

You can easily see why.

Answer 12

\(v_p\) : exponent of the prime \(p\) in the prime factorization of given number

Answer 13

Basically if you separate a number into a few parts such that the product is the number, then the individual valuations must add to the valuation of a prime in the given number.\[n = 2^5\cdot 3^7 = 2^2\cdot 3^7 \times 2^2 \times 2\]

Answer 14

You're only counting the number of 2s, haha. They must all add up to the number of 2s in the original number.

I feel like I am over-explaining it now.

Answer 15

Well, yes, but how does the formula work? Or is it just a fancy way to write everything with letters and doesn't help anything at all?

Answer 16

Because I understand what it does (how to count the powers), but I can't seem to get what is written with those symbols of product and summation.

Answer 17

It helps because you've broken down one big number into several smaller ones.
That's also how you do it in your mind... say you have to find the factorisation of 210 - then you first write 21 * 10 and then you decompose that into 7*3 * 2*5.
So decomposition always helps.

Answer 18

You're right. It doesn't help in simplifying anything. Just notation. But you should be comfortable with interpreting these by now. You do know what sum and product symbols mean, right ?

Answer 19

Yes, I do know what they mean.
What is bugging me is how \(v_p(k^k)\) becomes \(v_p(k)\cdot k\)

Answer 20

I'm talking about the fancy functions and product, sum symbola. Not the actual method

Answer 21

\[v_p(ab) = v_p(a) + v_p(b)\]You get this, right? Can you use that to prove \(v_p(k^k) = kv_p (k)\)?

Answer 22

Let me give you an example

Answer 23

what's the exponent of 2 in the prime factorization of 8^8 ?

Answer 24

Oh! I understood it now.
Somehow I didn't take notice that \(v_p()\) is a function that needs to be executed on each of teh number.
@ganeshie8 8.

Thank you for your help, @ParthKohli @ganeshie8 !

Answer 25

Oh, not 8.
for 2 it would be 24

Answer 26

So, \(v_2(8^8)= v_2(8) \cdot 8 = 3 \cdot 8 \)

Answer 27

Awesome! Also we should never ignore/neglect fancy notation. It really makes your work compact and neat sometimes. Have you heard of Leibniz and newton story about the symbol dy/dx ?

Answer 28

I haven't heard about the story, but I am quite familiar with the notation, going to look into it now (what would be the best way to search about information of it?)

Answer 29

I got that story from some calculus lecture. Gimme 2 minutes to pull it up

Answer 30

Its prof jerison :

Now, this is kind of a vague notion, this little bit here being an infinitesimal. It's sort of like an infinitely small quantity. And Leibniz perfected the idea of dealing with these intuitively. And subsequently, mathematicians use them all the time. They're way more effective than the notation that Newton used. You might think that notations are a small matter, but they allow you to think much faster, sometimes. When you have the right names and the right symbols for everything. And in this case it made it very big difference. Leibniz's notation was adopted on the continent and Newton dominated in Britain and, as a result, the British fell behind by one or two hundred years in the development of calculus. It was really a serious matter. So it's really well worth your while to get used to this idea of ratios. And it comes up all over the place, both in this class and also in multivariable calculus. It's used in many contexts.

Answer 31

First 5 mnts
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-15-antiderivatives/

Answer 32

Loved it! Thanks!