no. of zeros

[jagdish]

Find The no. of Zero at the end of ${^{2002}}C_{1001}$

[Babai]

Hi,I am a new member from West Bengal.Here is the solution:

We have to find $n\forall$ naturals such that $10^n\mid \dbinom{2002}{1001}$
$10=2*5$
The power of $2$ in $2002!$ is ${\lfloor\frac{ 2002} {2}\rfloor} +\lfloor\frac{ 2002}{2^2}\rfloor +.....+\lfloor\frac{ 2002}{2^{10}}\rfloor =1997$
The power of $5$ in $2002!$ is ${\lfloor\frac{ 2002} {5}\rfloor} +\lfloor\frac{ 2002}{5^2}\rfloor +.....+\lfloor\frac{ 2002}{5^{4}}\rfloor =499$

So,the power of $10$ in $2002!$ is min $(1997,499)=499$
According to the same manner we get $2^{992}\mid 1001!$ and $5^{249}\mid 1001!$
So,the power of $10$ in $1001!$ is min $(992,249) = 249$
As,$\dbinom{2002}{1001}=(2002!)\mid(1001!)^2$ so the power Of $10$ in $\dbinom{2002}{1001}$ is $499 -2*249=1$
So,number of zero at the end of $\dbinom{2002}{1001}$ =1

[Moon]

Hi, Babal. Welcome to the forum.

You might want to introduce yourself in the "introduction" sub-forum.

[jagdish]

Thanks babai