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# Help!

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Let $f(x) = x^2 - x + 2010$. What is the greatest common divisor of $f(100)$ and $f(101)$?

Jan 18, 2021

#1
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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  $$\boxed{10}$$

Jan 19, 2021
#2
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I just realized i made a typo, because x^2-x = x(x-1), but the answer remains the same

Jan 19, 2021
#3
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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.

Jan 19, 2021