Factorial Power

We need to know the prime factors of 15; they are 3 and 5. Since 3 would appear more occasionally in 128!, then all we have to calculate is the ${ 5 }^{ n }$ that appears in 128!. We can use Legendre's theory help on this. $\sum _{ n=1 }^{ \infty }{ \left\lfloor \frac { 128 }{ { 5 }^{ n } } \right\rfloor } \quad =\quad \left\lfloor \frac { 128 }{ 5 } \right\rfloor \quad +\quad \left\lfloor \frac { 128 }{ 25 } \right\rfloor \quad +\quad \left\lfloor \frac { 128 }{ 125 } \right\rfloor \quad +\quad \left\lfloor \frac { 128 }{ 625 } \right\rfloor \quad +\quad ...$ $= 25 + 5 + 1 + 0 + .... = 31$

The problem has been reported can anyone tell me why..??

As 15=3x5,we need to find the number of integers within and including 1-128 first.But fortunately, we can eliminate multiples of 3 from our consideration as there are many more of them than multiples of 5 within the limit. Now, 128/5=25.6,which implies that there are 25 multiples of 5 in [1,128]. But wait! Does only consideration of multiples of 5 suffice?
No.lets consider 75. 75=3x5x5. So,75 provides 2 fives,not just one. In fact any multiple of 25 will provide 2 fives in the factorized form of 128!. And to add a little more twist to the story, 125 ptovides 3fives. As 128/25=5.12, there are 5 multiples of 25 who provide 5 extra fives.125 itself provides an extra five. So, total number of fives in the factorized form of 128! is equal to 25+5+1=31. Answer is 31 :)