what is the largest possible value of N?

Question

I was told that the following problem can be solved using this method, but I don't understand why N equals the sum of the $4$ terms. Can anyone explain it to me? I guess that my problem is that I may not understand the meaning of "$2^{N}$ is a factor of $20!$". $\newcommand{\flr}[1]{\left\lfloor\,{#1}\,\right\rfloor}$ $$ N = \flr{20 \over 2}+ \flr{20 \over 2^{2}} +\flr{20 \over 2^{3}} + \flr{20 \over 2^{4}} = 10 + 5 + 2 + 1 = 18 $$

If $2^{N}$ is a factor of $20!$, what is the largest possible value of $N$ ?.

Answer 1

$2^N$ is a pure power of $2$, $20!$ is not. So, while dividing $20!$ by $2^N$, think of it as diving it $N$ times by $2$. Now, in order for $2^N$ to be a divisor of $20!$, there must be at least $N$ many $2$'s in its factorization. So, we count how many $2$'s are there. That'll be the maximum possible value of $N$.

As for the calculation, note that $20!=1 \cdot 2 \cdots 20$. How many multiples of $2$ are there? Exactly $\lfloor {20 \over 2} \rfloor$. Here comes the tricky (well, a bit) part. The multiples of $4$ contribute two $2$'s. So, we count them once again. Continuing, we count the number of multiples of $2^n$ under $20$, that is $\lfloor {20 \over 2^n} \rfloor$ . In this case, after $2^4$, we get zero number of multiples. Hence the solution is $10+5+2+1=18$.

Answer 2

$$20!=1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot16\cdot17\cdot18\cdot19\cdot20.$$

Now strike out every odd factor

$$2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16\cdot18\cdot20$$

and divide all these $10$ factors by $2$. You obtain

$$1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10.$$

This shows that the multiplicity of $2$ in $20!$ is $20/2$ plus the multiplicity in $10!$. For the same reason, the multiplicity in $10$ is $10/2$ plus the multiplicity in $5!$. And do on.