The natural question one would ask is what the difficulty with the topic is that you are facing.
I followed along until the third step, where it was explained that there are 514 twos in 517, but I am having trouble understanding the fourth step.
514 twos in 517? Where are you seeing that?
The third step
See the previous video. In other words, what is the greatest integer t such that \frac{517!}{2^t} is an integer?
517! = 1 * 2 * 3 * 4 * 5… * 516 * 517
Hence, 517! will have the following multiples of 2…
2, 4, 6, 8, 10…, 516
Which you can rewrite as,
2*1, 2*2, 2*3, 2*4, 2*5…, 2*258
Now the problem is, each one can have multiple 2s, Hence better way to see it like…
Let’s remove all 2s, then new series will be like,
1, 2, 3, 4, 5…, 258 and we have taken out 258 2’s from this (1, 2*1, 3, 2*2, 5, 2*3…, 257, 2*129)
Let’s remove all 2s again, and new series will be like,
1, 2, …, 129 further taking out 129 2’s from this. (Note, it’s just integer part of 258/2 = 129)
We will keep on doing this until we are left with no more 2’s.
and then we will add all of them. And it will look like this,
517/2 = 258.5 => 258 (integer part)
258/2 = 129
129/2 = 64.5 => 64 (integer part)
64/2 = 32
32/2 = 16
16/2 = 8
8/2 = 4… 4/2 = 2… 2/2 = 1… 1/2 =0.5 =>0
Now let’s add all 2s.
258 + 129 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 514
In short: Quick steps will look like this,
(517/2) + (517/4) + (517/8) + (517/16) … + (517/516)
Notice that above we are diving 258 by 2, and we got 258 by dividing 517 by 2. So it’s ok write 258/2 or 517/4.
So 517! will have 514 2s. Hope that may help clear out basics. 