Trailing Zeros

Hi Samyak 

25! ( 1 + 26 + 26 × 27 + 26 × 27 × 28 + 26 × 27 × 28 × 29 × 30 ) 

25! ( 1 + 26 + 26 × 27 + 26 × 27 × 28 + 26 × 27 × 28 × 29 × 30 )

25! has 6 trailing zeros and, the term inside the bracket is divisible by 5 

Hence, 6 +1  , 7 trailing zeros . 

Highest power of 2 in 25! = [25/2] + [25/2^2] + [25/2^3] +........

12 + 6 + 3 + 1 = 22 

So , the highest power of 8 in 25! = [22/3] = 7 

@neha-singh 

@neha-singh 

(800)!

number of trailing zeroes = 800/5 + 160/5 + 32/5 + 6/5

= 160 + 32 + 6 + 1 = 199

@mayanksri 

Please find the solution below.

30! = 2^26 x 3^14 x 5^7 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1

Now the for the first non zero digits,lets first remove the terms which are causing the trail of zeroes at the end...which is 5^7 x 2^7....so u r left with....

2^19 x 3^14 x 7^4 x 11^2 x 13^2 x 17^1 x 19^1 x 23^1 x 29^1

the last digit of this pdt will be 8 x 9 x 1 x 1 x 9 x 7 x 9 x 3 x9 = 8..
so the first non zero digit from the right will be 8.

@mayanksri 

power of 5 in 100!*200!=73. Thus number of zeros at the end of this num = 73.

@anna 

Power of 5 in 1!*2!*3!*4!*5*................................*50! = 262. Thus number of zeros at the end of this num = 262.

@mayanksri 

Please find the solution below,

N = 4*8*12*....*1000 = (4*1)*(4*2)*(4*3)*.....*(4*250) = 4²⁵⁰*(1*2*3*...250) = 2⁵⁰⁰*250!
Now 250! = 2²⁴⁴*3¹²³*5⁶²*7⁴⁰*11²⁴*13²⁰*17¹⁴*19¹³*23¹⁰*29⁸*31⁸*37⁶*......
Hence the number of zeroes at the end of N is 62.

@mayanksri