Coefficients Party, Part 6
How many of the coefficients in the expansion of ( x − 1 ) 2 0 1 1 − ( x 2 0 1 1 − 1 ) are divisible by 2 0 1 1 ?
The answer is 2010.
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2 solutions
Do you know the application of this problem? Something very interesting regarding k results if every term in the expansion of ( x − 1 ) k − ( x k − 1 ) is divisible by k .
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That means ( x − 1 ) k ≡ x k − 1 ( m o d p ) for prime p ! Sample problem: how many of ( 1 1 0 0 ) , ( 2 1 0 0 ) , ( 3 1 0 0 ) , … , ( 1 0 0 1 0 0 ) are divisible by 1 0 0 ?
(x-1)2011 will have 2012 terms.The first term being x^2011 and the last term,-1.Both of them will get cancelled as you can see in the question.Now after doing some simple calculation,with the help of binomial theorem,we can find that all the terms except the first and the last ones in the expansion of (x+2011)^2 are divisible by 2011.Thus 2010 terms are divisble by 2011.
The expression when simplified will cancel x 2 0 1 1 and the − 1 in the expansion of ( x − 1 ) 2 0 1 1 leaving ( 2 0 1 1 + 1 − 1 − 1 ) = 2 0 1 0 terms. All of these terms' absolute value will be of the form ( k 2 0 1 1 ) x k where k is a positive integer in range [ 1 , 2 0 1 0 ] . Also since 2 0 1 1 is a prime, 2 0 1 1 ∣ ( k 2 0 1 1 ) for 1 ≤ k < 2 0 1 1 . Hence all these terms are divisible by 2 0 1 1 . The answer is 2 0 1 0 .