16. All the positive integers which are co-prime to 2012 are grouped in an increasing order in such a way that the $n^{\text {th }}$ group has $2 n-1$ numbers. So, the first three groups in this grouping are (1), $(3,5,7),(9,11,13,15,17)$. It is known that 2013 belongs to the $k^{\text {th }}$ group. Find the value of $k$.
(Note: Two integers are said to be co-prime if their greatest common divisor is 1.)
16. Answer: 32

Solution. Note that $2012=2^{2} \times 503$, and that 503 is a prime number. There are 1006 multiples of 2 less than or equal to 2012 ; there are 4 multiples of 503 less than or equal to 2012; there are 2 multiples of 1006 less than or equal to 2012. By the Principle of Inclusion and Exclusion, there are $1006+4-2=1008$ positive integers not more than 2012 which are not co-prime to 2012 . Hence there are $2012-1008=1004$ positive integers less than 2012 which are co-prime with 2012 . Thus, 2013 is the $1005^{\text {th }}$ number co-prime with 2012. Note also that the sum of the first $n$ odd numbers equals $n_{2}^{2}$, and that $31^{2}<1005<32^{2}$, the number 2013 must be in the 326 th group. Hence $k=32$.