Elementary number theory

[river osprey]

Hello, I got a short lecture on number theory as an introductory lesson to university Algebra 1 (I assume that's abstract algebra in the US).
I am taking this course after Linear Algebra, so I have some knowledge from there.

What I have question about is a rather simple task:
"Determine with how many zeroes does $144!$ end with." ('!' being factorial sign ex. $5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$)

It wasn't explained well, what are we looking for? All I see done is choosing number 10 and then 2 and 5 (because they're primes and factors of 10).
$[\frac{100}{2^1}] + [\frac{100}{2^2}] + [\frac{100}{2^3}] + [\frac{100}{2^4}] + [\frac{100}{2^5}] + [\frac{100}{2^6}] + [\frac{100}{2^7}] + ...$ (last term as well as all the rest count as 0)
And all this equals to 77.
Now doing similarly with 5:
$[\frac{100}{5^1}] + [\frac{100}{5^2}] + [\frac{100}{5^3}] + ...$ )(last term as well as all the rest count as 0)
And all this equals to 24.

Now, $144! = 2^24\cdot 5^77\cdot something...$

And conclusion from here is that $144!$ ends with 24 zeroes but I don't understand how we got to here? Can anybody explain steps in detail and why they're performed?

[remote ridge]

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[fringe cypress]

danilojonic

[floral elbow]

yes surely

The exponent of the prime p in n! is given by,

$\sum_{k=1}^{\infty}\lfloor\frac{n}{p^k}\rfloor$

[fringe cypress]

Cofailure

[floral elbow]

here is how we come up with this

first note that the number of multiples of m below n is floor(n/m)

The first term accounts for all the numbers with a prime power of atleast 1

the second term accounts for the numbera with a prime power of atleast 2

.

.

.

the k-th term accounts for the numbers with a prime power of atleast k

notice how the numbers with the prime power of m are counted exactly m times, once in the j-th term for every 1=< j =< m

Now, if a number has the prime powers of 5 and 2 as X and Y respectively

we can see exactly min{X, Y} zeroes at the end as we group 5s and 2s to make tens!

@river osprey

[river osprey]

I didnt get this part

doesnt seem intuitive at all

n is the factorial number?

p is then some prime divisor

but k and floor are complicated to understand here

[jovial vault]

consider that 414! has 82 numbers that are divisible by 5

however, 16 of those are divisible by 25

and 3 by 125

[floral elbow]

did you read what i wrote?

okay here's another idea it'll make you understand surely

do you know how to code?

[river osprey]

yes in C

what confused me is the general approach

can you give another realistic example and explain on that one in detail?

[floral elbow]

we try it for p=3

take n=500

make columns

k | divisible by 3? | divisible by 9? | divisible by 27?...

then fill out each 1<=k<=n=500 in the left most column

and look at the table

[river osprey]

can I also use euclidean algorithm for this?

I recall that being done as well

a = q*b + r for calculating gcd, where a is the larger number, b is the smaller, q is the multiplier of b for how many times b fits into a and r is the remainder

maybe fundamental theorem of arithmetic is also being used here? That states that every natural n (n>=2) can be rewritten as a product of prime degrees, so
$n = p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_k^{a_k}$ where $p_1, p_2,...,p_k$ are primes

[fringe cypress]

danilojonic

[fresh whale]

@river osprey

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[young talon]

+close