Factorial remainder!

We count the number of $2$ 's in the unique prime factorization of $33!$ . We do this using the fact that $2$ is a prime and using De-Polignac's formula .

Let $s$ denote the maximum power of $2$ that divides that $33!$ . Then, we have,

$s=\left\lfloor\frac{33}{2}\right\rfloor+\left\lfloor\frac{33}{4}\right\rfloor+\left\lfloor\frac{33}{8}\right\rfloor+\left\lfloor\frac{33}{16}\right\rfloor+\left\lfloor\frac{33}{32}\right\rfloor=16+8+4+2+1=31$

Since $s$ is the maximum power of $2$ that divides $33!$ and $s=31$ , we have,

$2^s=2^{31}|33!$

$\therefore\quad 2^{31}|33!\iff 33!\equiv 0\pmod{2^{31}}\iff 33!\bmod 2^{31}=0$

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Note: $x|y$ denotes " $x$ divides $y$ ".

We have , 33!=1 * 2 * 3*...... * 33 , Now the only even prime factor 2 comes in 4,6,8,...,and 32. That's 2 * 2^2 * 2 * 2^3.... * 2^5 =2^31. Hence 33! is completely divisible by 2^31. So 33!mod2^31=0.

33!=33 x 32 x 31 x 30 x ...... x 2. 2= 2. 4= 2 x 2. 6= 2 x 3. ...=... . 32= 2 x 2 x 2 x 2 x 2.

Then, count how many "2". If the 2 more or = 31, it means the answer is 0 because that is the factor of 33!