SOLUTION: The number N = 111….1 consists of 2010 ones. It is exactly divisible by 3. How many zeroes are there in the quotient of N/3

Notice that


    111 : 3 = 37;


    111111 : 3 = 37037;


    111111111 : 3 = 37037037,


    and so on :  each group consisting of 3 digits, generate 037 in the quotient, when is divided by 3.



The given number  111...1 consists of 2010 ones.

So, there are  2010 : 3 = 670 groups each consisting of 3 ones, like 111.



Each such group produces one zero, and we should throw out the very first such zero, since we don't want 

to have zero as the leading digit.



Then (and thus) we have 670 - 1 = 669 zeroes in the quotient.     ANSWER