Hot Meatballs 1

The highest power of $2=p$ (say) such that $2^p$ divides $2000!$ is given by

$p=\lfloor \frac{2000}{2}\rfloor +\lfloor \frac{2000}{2^2}\rfloor +\lfloor \frac{2000}{2^3}\rfloor +...$

$=1000+500+250+125+62+31+15+7+3+1+0=\boxed {1994}$ .

Well we all knew that $2000!$ = $1 × 2 × 3 × 4 × 5 ... 1999 × 2000$

That means: $K_2$ (2000!)= $K_2$ (1) + $K_2$ (2) + $K_2$ (3) + $K_2$ (4) ...... $K_2$ (1999) + $K_2$ (2000) = $\displaystyle\sum_{n=1}^{2000}$ ( $K_2$ (n))

• $\color{gold}{\textbf{The Main 2 Rules We will use}}$

$\boxed{1}$ $\displaystyle\sum_{i=1}^{n^{m}×X}(K_n(i))$ = $\displaystyle\sum_{i=1}^{X}(K_n(i))$ + ( $\frac{(n^m-1)×X}{n-1}$ )

$\boxed{2}$ $\displaystyle\sum_{i=1}^{X}(K_n(i))$ = $X-B(X)$ $\rightarrow$ (Which $B(X)$ is the sum of all one digits in binary of X )

By Using Rule 1

• $\displaystyle\sum_{n=1}^{2000}$ ( $K_2$ (n)) = $\displaystyle\sum_{n=11}^{125×2^{4}}$ ( $K_2$ (n)) = $\displaystyle\sum_{n=1}^{125}$ ( $K_2$ (n)) + $125$ × $\dfrac{2^{4}-1}{2-1}$

By Using Rule 2

• $\displaystyle\sum_{n=1}^{125}$ ( $K_2$ (n)) + $125$ × $\dfrac{2^{4}-1}{2-1}$ = $125$ - $B(125)$ + $(15×125)$ = $125$ - $B(125)$ + $1875$

$B(125)$ = $B(124)$ + $1$ = $B(31$ 0 + $1$ = $B(2^{5}-1$ ) + $1$

A cool fact $B(2^{n}-1$ ) = n

• $B(2^{5}-1$ ) + $1$ = $(5)$ + $1$ = $\boxed{6}$

That means $\displaystyle\sum_{n=1}^{2000}$ ( $K_2$ (n)) $= 125-6+1875 = \boxed{1994}$

$\fbox{\color{#BBBBBB}{Another Solution:}}$

By Using Rule 2

• $\displaystyle\sum_{n=1}^{2000}$ ( $K_2$ (n)) = $2000-B(2000)$

$B(2000)=B(2^{5}×125)$ you can remove the 2 factor because it is only changing digits place without adding any digit

$B(2000)=B(2^{5}×125)=B(125)=6$

So $\displaystyle\sum_{n=1}^{2000}$ ( $K_2$ (n)) = $2000-B(2000)$ = $2000-6 = \boxed{1994}$

$\fbox{\color{#D61F06}{History Fact :)}}$

This date when the colombian football player (Andrés Escobar) killed by mafia for his mistake which made Colombia out from Fifa World Cup 1994