All greatest common divisors of (M, 2010) are multiples of 67. Since M is a multiple of both 271 and 67 you know that , , so what do you think about x?
Suppose a certain studentís ID number M satisfies
gcd(M, 2010) > gcd(M, 271) > 1.
Find all possible values for gcd(M, 2010). Be sure to explain your reasoning. [Note: both 271 and 67 are prime.]
I have gotten most of the way through the question....but will wait to see if anyone else has any insights. It is very possible that I am stuck because one of my steps lead me in the wrong direction...