Number System

Find the number of zeros at the end of 36!^36!............

Regards,

Dipanjan

__________________

n/a

Number of zeros at the end
Number of zeros at the end of 36! = [36/5] + [36/25] = 7 + 1 = 8

So Number of zeros at the end of 36! 36! =  8 36!

 

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n/a

RE

I know urs way, bt my qs is how can u decide dt that process gives the no. of zeros at the end, nt overall.....

Watin 4 re......

Thank you,

Dipanjan

Sorry DipanjanI am not
Sorry Dipanjan
I am not getting your question
Number of Zeros depends upon number of 5’s and 2’s and that can be calculated by
Number of 5’s in n! = [n/5] + [n/52] + [n/53] + . . . .
__________________

n/a

NIILM

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@Praveen

My qs is

No. of zeros in n!=[n/5]+[n/5^2]+......

Bt ths procees can calculate the total no. of zeros in n!.I thnk by this process we cant calculate the no. of zeros at the end of 36!.......Plz look closely on d qs.........What do u thnk is it sufficient to cal d no. of zeros at d end of 36!^36!...

If u thnk dt process is correct thn plz explain it..............

Thank You.

Dipanjan

my..ans.

its 36!*8

(100)^3 has 6 zeros

i.e num of 0's in the base *the power

36! has 8 zeros..

so (36!)^36! has...8*36! zeross....

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