Function can be tricky (4)
Let f ( x ) be a quadratic polynomial with integer coefficients. If the value of f ( 0 ) , f ( 3 ) and f ( 4 ) are pairwise distinct but all belong to the set { 7 , 7 1 , 7 1 0 , 7 1 0 2 } , find the sum of all possible values of f ( 1 ) .
The answer is 7890.
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1 solution
(playing devil's advocate) Why can the first step be done? IE Can all quadratic polynomials with integer coefficients be expressed in that way for integer A, B, and C?
Note: The conclude that a − b ∣ f ( a ) − f ( b ) can be obtained directly / is a "well known" result.
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The first is "yes", since it is a matter of finding quotients and remainders when dividing by a two monic polynomials, the first quadratic and the second linear.
That a − b divides f ( a ) − f ( b ) is a bit fiddly in general (you gave to concatenate the fact that a − b divides a n − b n for all n ), but the fact that a divides f ( a ) − f ( 0 ) , which is all I use, is trivial. That yields a nicer approach.
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Indeed! I wanted to point out that it's an application of the remainder factor theorem with a slight twist. We first find the remainder when divided by x − 3 , and then the remainder of the quotient when divided by x − 4 .
Most of the time, people are used to dividing out by the same term repeatedly, which leads to (say) f ( x ) = A ( x − 3 ) 2 + B ( x − 3 ) + C , which is also the maclaurin series expansion :)
Interestingly, 7 + 7 1 + 7 1 0 + 7 1 0 2 = 7 8 9 0 . I found this by using a flawed method, but I still managed to get the right answer. :P
We can write f ( x ) = A ( x − 3 ) ( x − 4 ) + B ( x − 3 ) + C where A , B , C are integers, and f ( 0 ) = 1 2 A − 3 B + C f ( 3 ) = C f ( 4 ) = B + C In particular, f ( 0 ) − f ( 3 ) = 3 ( 4 A − B ) is a multiple of 3 and f ( 0 ) − f ( 4 ) = 4 ( 3 A − B ) is a multiple of 4 . Thus there are only four possibilities: f ( 0 ) 7 1 7 1 0 7 7 1 0 2 f ( 3 ) 7 1 0 7 1 7 1 0 2 7 f ( 4 ) 7 7 1 0 2 7 1 7 1 0 A − 2 2 9 1 8 1 1 − 2 3 4 9 7 6 7 B − 7 0 3 7 0 3 1 − 7 0 3 1 7 0 3 C 7 1 0 7 1 7 1 0 2 7 f ( 1 ) 7 4 2 − 3 1 2 5 7 0 7 0 3 2 0 3 and hence the answer is 7 4 2 − 3 1 2 5 + 7 0 7 0 + 3 2 0 3 = 7 8 9 0 .