no. of zeros

[man111]

find no. of zero at the end of [tex]\displaystyle {^{2002}}C_{1001}[/tex]

[martin123456]

what is [tex]\displaystyle {^{2002}}C_{1001}[/tex] ?

[mkmarinov]

I suppose it is [tex]\frac{2002!}{(1001!)^2}[/tex]

[martin123456]

The degree of 5 in 2002! is deg5 =
[tex][\frac{2002}{5}]+\frac{2002}{5^2}]+\frac{2002}{5^3}]+[\frac{2002}{5^4}]=400+80+16+3=499[/tex].
The degree of 2 in 2002! is deg2 =
[tex][\frac{2002}{2}]+[\frac{2002}{2^2}]+\frac{2002}{2^3}]+[\frac{2002}{2^4}]+[\frac{2002}{2^5}]+[\frac{2002}{2^6}]+[\frac{2002}{2^7}]+[\frac{2002}{2^8}]+[\frac{2002}{2^9}]+[\frac{2002}{2^{10}}][/tex][tex]=1001+500+250+125+62+31+15+7+3+1=1995[/tex]. So [tex]2002!=2^{1995}5^{499}x[/tex], [tex](x,10)=1[/tex].

The degree of 5 in 1001! is deg5 =
[tex][\frac{1001}{5}]+\frac{1001}{5^2}]+\frac{1001}{5^3}]+[\frac{1001}{5^4}]=200+40+8+1=249[/tex].
The degree of 2 in 2002! is deg2 =
[tex][\frac{1001}{2}]+[\frac{1001}{2^2}]+\frac{1001}{2^3}]+[\frac{1001}{2^4}]+[\frac{1001}{2^5}]+[\frac{1001}{2^6}]+[\frac{1001}{2^7}]+[\frac{1001}{2^8}]+[\frac{1001}{2^9}][/tex][tex]=500+250+125+62+31+15+7+3+1=994[/tex]. So [tex](1001!)^2=2^{1988}5^{498}y[/tex], [tex](y,10)=1[/tex].

So [tex]\frac{2002!}{(1001!)^2}=2^{7}5z[/tex], [tex](z,10)=1[/tex], so is divisible by 10

[man111]

thanks martin got it.