Factoring Factorial Part 1

First off, let's rewrite $10!$ with its prime factors. $$ 1) Calculate ${ 2 }^{ A }$ $\sum _{ A=1 }^{ \infty }{ \left\lfloor \frac { 10 }{ { 2 }^{ A } } \right\rfloor } \quad =\quad \left\lfloor \frac { 10 }{ 2 } \right\rfloor \quad +\quad \left\lfloor \frac { 10 }{ 4 } \right\rfloor \quad +\quad \left\lfloor \frac { 10 }{ 8 } \right\rfloor \quad +\quad ...\quad =\quad 5\quad +\quad 2\quad +\quad 1\quad +\quad ...\quad =\quad 8$ 2) Calculate ${ 3 }^{ B }$ $\sum _{ B=1 }^{ \infty }{ \left\lfloor \frac { 10 }{ { 3 }^{ B } } \right\rfloor } \quad =\quad \left\lfloor \frac { 10 }{ 3 } \right\rfloor \quad +\quad \left\lfloor \frac { 10 }{ 9 } \right\rfloor \quad +\quad ...\quad =\quad 3\quad +\quad 1\quad +\quad ...\quad =\quad 4$ 3) Calculate ${ 5 }^{ C }$ $\sum _{ C=1 }^{ \infty }{ \left\lfloor \frac { 10 }{ { 5 }^{ C } } \right\rfloor } \quad =\quad \left\lfloor \frac { 10 }{ 5 } \right\rfloor \quad +\quad ...\quad =\quad 2\quad +\quad ...\quad =\quad 2$ 4) Calculate ${ 7 }^{ D }$ $\sum _{ D=1 }^{ \infty }{ \left\lfloor \frac { 10 }{ { 7 }^{ D } } \right\rfloor } \quad =\quad \left\lfloor \frac { 10 }{ 7 } \right\rfloor \quad +\quad ...\quad =\quad 1\quad +\quad ...\quad =\quad 1$ Therefore, $10!\quad =\quad { 2 }^{ A }\quad \times \quad { 3 }^{ B }\quad \times \quad { 5 }^{ C }\quad \times \quad { 7 }^{ D }\quad =\quad { 2 }^{ 8 }\quad \times \quad { 3 }^{ 4 }\quad \times \quad { 5 }^{ 2 }\quad \times \quad { 7 }^{ 1 }$ Now, let's rewrite $10!$ as what the question asks for. 3 will stay as 3, there fore $a = 4$ , 2 and 5 will create 10 since 5 does not go into another exponent, therefore $b = 2$ , 2 and 7 will create 14 since 7 does not go into another exponent, therefore $c = 1$ , the rest is ${ 2 }^{ 5 }$ or 32, therefore $d = 1$ . $d\sqrt { abc } \quad =\quad 1\sqrt { 4\times 2\times 1 } \quad =\quad \sqrt { 8 } \quad =\quad 2\sqrt { 2 }$