# $2007^{201} + 2019^{201} - 1982^{201} - 2044^{201}$ is divisible by which of the following?

I reduced it to

$$(2013-6)^{201} + (2013+6)^{201} - (2013-31)^{201} - (2013+31)^{201}$$

But this way I only get option D. How do I check for the other options. (Multiple correct question)

• Factorise 2013. What do you get? Jun 10 '20 at 6:56
• 25 is definitely a factor Jun 10 '20 at 6:59
• $2019=1982+37$ and $2044=2007+37$. Jun 10 '20 at 7:03
• Do you know Fermat's little theorem? The Euler extension of this? Can you deal with divisibility by the small primes involved - $2, 3, 5^2$? What of the larger primes? Jun 10 '20 at 7:57
• I know how to deal with divisibility of high powers by small numbers using binomial theorem. Jun 10 '20 at 11:58

$$2007^{201}+2019^{201}$$ is divisible by $$2007+2019=4026=2\cdot2013$$ and
$$1982^{201}+2044^{201}$$ is divisible by $$1982+2044=4026=2\cdot2013$$.
Thus, our expression is divisible by $$2013$$.
The expression is an even number, $$2019-1982=37$$ and $$2007-2044=-37$$, which says that our expression is divisible by $$74$$.
Also, $$2007-1982=-25$$, $$2019-2044=-25$$, which says that our expression is divisible by $$50$$ and by $$50\cdot37=1850.$$