2
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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)

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5
  • 1
    $\begingroup$ Factorise 2013. What do you get? $\endgroup$ Jun 10 '20 at 6:56
  • $\begingroup$ 25 is definitely a factor $\endgroup$ Jun 10 '20 at 6:59
  • 1
    $\begingroup$ $2019=1982+37$ and $2044=2007+37$. $\endgroup$
    – mathlove
    Jun 10 '20 at 7:03
  • $\begingroup$ 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? $\endgroup$ Jun 10 '20 at 7:57
  • $\begingroup$ I know how to deal with divisibility of high powers by small numbers using binomial theorem. $\endgroup$ Jun 10 '20 at 11:58
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$2007^{201}+2019^{201}$ 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 $50\cdot37=1850.$

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