Ramaiah Problem

660

Not sure if I'm doing this correctly as I'm not well practiced with congruent arithmetic

6666......66666 for 2002 terms is the same as 2/3(10^2003-1)
Dividing by 2002 is 2*1001, and division of 2/3(10^2003-1) by 2 is trivial
1/3(10^2003-1)

now the problem is to divide 10^2003 by 1001,
lets look at 1001
1000=10^3=-1(mod1001)
10^2003=10^2001*10^2
(10^3)^667=10^2001
so (-1^667)(mod1001)=-1(mod1001)
10^2(mod1001)=100(mod1001)
-1(mod1001)*100(mod1001)=-100(mod1001)=901(mod1001)
so(10^2003-1)(mod1001)=(901-1)(mod1001)=900(mod1001)
1/3*900(mod1001)=300(mod1001)
doubling to get back to mod2002
300(mod1001)=600(mod2002)
So the residual class (remainder) to
666......6666=2/3(10^2003-1) divided by 2002 should be 600

Rap

2002*333 = 666666

We can lop off groups of six 6s from the left side of the 2002 6s with out affecting the remainder.

Since 6*333 = 1998, we can lop off 1998 6s leaving 6666.

6666 - 3*2002 = 660

300(mod1001)=600(mod2002)

Huh?

Markr that was elegant.

I am getting 660 as well:

If we call &n the n-digit number whose digits are all 6, we get the following periodic table:

n &n mod 2002
1 6
2 66
3 666
4 660
5 600
6 0
7 6
8 66
...
1998 0
1999 6
2000 66
2001 666
2002 660

/Kayyam

OK I see my error---Still learning

Rap