Help!

First, realize that \(x^2-x=x(x+1)\). Also, realize that the prime factorization of 2010 is \(2\cdot3\cdot5\cdot67\). Therefore we can rewrite f(100) and f(101) like this:

\(f(100)=100\cdot99+2\cdot3\cdot5\cdot67 = 10(10\cdot99+3\cdot67) \\f(101)=101\cdot100+2\cdot3\cdot5\cdot67=10(10\cdot101+3\cdot67)\)

Since we cannot factor further, the answer to this question is  \(\boxed{10}\)

I just realized i made a typo, because x^2-x = x(x-1), but the answer remains the same

Thanks textot  

Here is another method:

\(f(100)=100^2-100+2010\)

\(f(101)=101^2-101+2010\\ f(101)=(100+1)^2-101+2010\\ f(101)=100^2+200+1-101+2010\\ f(101)=100^2-100+200+2010\\ f(101)=f(100)+200\\\)

Now my problem has changed to, what is the highest common factor of

  \(100^2-100+2010 \qquad and \qquad 200\)

10 goes into both so I will divide by 10

\(1000-10+201 \qquad and \qquad 20\)

The last digit of the first one is 1.   Nothing ending in 1 will have any common factors with 20, (it would have to end in a multiple of 2 or 5)

So

The hight common factor is 10.