SOLUTION: What is the 2002 digit to the right of the decimal in the expansion of 1/13?

We do the long division:

    .076923076923··· 
  ---------------
13)1.000000000000···
     91
   ----
      90    
      78
      --
      120
      117   
      ---
        30
        26
        --  
         40
         39
         --  
          100
           91
          ---   
            
So you see that 1/13 = .076923076923076923076923

So it's a infinite repetition of the block of 6 digits

078923 

Now we divide 2002 by 6

   333 
 ----- 
6)2002
  18
  --
   20
   18
   --
    22
    18
    --
     4

So that's 333 complete blocks of 6 digits 076923, and the remainder
4 shows that the 2002nd digit is 4 digits into the 334th block. 333x6 
is 1998. So the 1999th digit starts the 334th block with the first 
digit of the block 0. The 2000th digit is the 2nd digit in the block, 
7. Then the 2001st digit is the 3rd digit in the block, or 6, and 
voila, the 2002nd digit is the 4th digit in the block, or 9.

Answer: 9.

Edwin