My problems #7

$f\left( n \right) =n\times \sum _{ i=0 }^{ n-1 }{ f\left( i \right) } \\ \qquad =n\quad \left[ f\left( 0 \right) +f\left( 1 \right) +\quad ...\quad +f\left( n-1 \right) \right] \\ \frac { f\left( n \right) }{ n } =\quad f\left( 0 \right) +f\left( 1 \right) +\quad ...\quad +f\left( n-1 \right) \\ f\left( n+1 \right) =\quad \left( n+1 \right) \sum _{ i=0 }^{ n }{ f\left( i \right) } \\ \qquad \qquad =\quad \left( n+1 \right) \left[ f\left( 0 \right) +f\left( 1 \right) +\quad ...\quad +f\left( n-1 \right) +f\left( n \right) \right] \\ \qquad \qquad =\quad \left( n+1 \right) \left( \frac { f\left( n \right) }{ n } +f\left( n \right) \right) \\ \qquad \qquad =\quad f\left( n \right) \left( \left( n+1 \right) \left( \frac { 1 }{ n } +1 \right) \right) \\ \qquad \qquad =\quad f\left( n \right) \left( \frac { { n }^{ 2 }+2n+1 }{ n } \right) =f\left( n \right) \quad \frac { { \left( n+1 \right) }^{ 2 } }{ n } \\ \therefore \quad f\left( n \right) =\frac { { n }^{ 2 } }{ n-1 } f\left( n-1 \right) \\ \qquad \qquad =\frac { { n }^{ 2 } }{ n-1 } \frac { { \left( n-1 \right) }^{ 2 } }{ n-2 } f\left( n-2 \right) \\ \qquad \qquad =\frac { { n }^{ 2 } }{ n-1 } \frac { { \left( n-1 \right) }^{ 2 } }{ n-2 } \quad ...\quad \frac { { 2 }^{ 2 } }{ 1 } f\left( 0 \right) \\ \qquad \qquad ={ n }^{ 2 }\left( n-1 \right) \left( n-2 \right) \quad ...\quad \left( 2 \right) \left( 1 \right) =n\quad .\quad n!\\ \therefore \quad f\left( 20 \right) =\quad 20\quad .\quad 20!\\ 20\quad =\quad { 2 }^{ 2 }.5\\ 20!\quad =\quad 20.19.18....2.1\quad it\quad has\quad 10\quad even\quad numbers\quad ({ 2 }^{ 10 })\\ those\quad 10\quad even\quad after\quad dividing\quad by\quad 2\quad will\quad be\quad 5\quad even\quad numbers({ 2 }^{ 5 })\\ and\quad those\quad 5\quad will\quad be\quad 2\quad even\quad numbers({ 2 }^{ 2 })\quad and\quad \quad then\quad 1\quad even\quad (2)\\ p=10+5+2+1=18\\ this\quad given\quad by\quad p=\sum _{ i=1 }^{ \infty }{ \left\lfloor \frac { 20 }{ { 2 }^{ i } } \right\rfloor } =18\\ \therefore \quad { m }^{ 2 }=\quad { 2 }^{ 18\quad +\quad 2 }\quad =\quad { 2 }^{ 20 }\quad \Rightarrow \quad \boxed { m=\quad { 2 }^{ 10 }=\quad 1024 }$

The following are the values of $f(n)$ for $0 \le n \le 20$ (I did it with a spreadsheet).

$\begin{array} {rrrcrrr} n & f(n) & \dfrac {f(n)}{n!} &|& n & f(n) & \dfrac {f(n)}{n!} \\ \hline \\ 0 & 1 & 1 &|& 11 & 439084800 & 11 \\ 1 & 1 & 1 &|& 12 & 5748019200 & 12 \\ 2 & 4 & 2 &|& 13 & 80951270400 & 13 \\ 3 & 18 & 3 &|& 14 & 1220496076800 & 14 \\ 4 & 96 & 4 &|& 15 & 19615115520000 & 15 \\ 5 & 600 & 5 &|& 16 & 334764638208000 & 16 \\ 6 & 4320 & 6 &|& 17 & 6046686277632000 & 17 \\ 7 & 35280 & 7 &|& 18 & 115242726703104000 & 18 \\ 8 & 322560 & 8 &|& 19 & 2311256907767810000 & 19 \\ 9 & 3265920 & 9 &|& 20 & 48658040163532800000 & 20 \\ 10 & 36288000 & 10 &|& & & \end{array}$

It can be seen that $f(n) = n\dot{}n!\quad \Rightarrow f(20) = 20\dot{} 20!$

The highest power-of-2 divisor of $20$ is $4=2^2$ ,

The exponent of the highest power-of-2 divisor of $20!$ is given by:

$\displaystyle \begin{aligned} p & = \sum _{i=1} ^\infty {\lfloor \frac {20}{2^i} \rfloor} \\ & = \lfloor \frac {20}{2} \rfloor + \lfloor \frac {20}{r4} \rfloor + \lfloor \frac {20}{8} \rfloor + \lfloor \frac {20}{16} \rfloor \\ & = 10 + 5 + 2 + 1 \\ & = 18 \end{aligned}$

Therefore, the largest power-of-2 divisor of $f(20)$ is: $2^2\dot{}2^{18} = 2^{20} = \left( 2^{10} \right)^2$

$\Rightarrow m = 2^{10} = \boxed{1024}$