Seeing 2020
Let f be a polynomial function with integer coefficients such that f ( 2 0 1 7 ) = 1 . Which of the following can equal f ( 2 0 2 0 ) ?
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6 solutions
Inspired solution! I think there is correction on the second line of f (a)-f (b), c0 must be vanished
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Thanks for spotting that! I edited the solution.
What if f(x) has real coefficients? Is this method can also be used too?
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Unfortunately no. If a 1 , a 2 , . . . , a n and b 1 , b 2 , . . . , b n be 2 n real numbers such that the a i are pairwise distinct and such that f ( a i ) = b i for all i , then I can always find a polynomial (in fact infinitely many polynomials) with real coefficients such that the above relation(s) hold true. See Lagrange Interpolation for such a construction.
I don't know how I saw 2014 among options...End up by selecting at least two options..
Nice, tidy way of putting it all together. I'm just the tiniest bit unsettled by the bit after "Going back to the problem, we have" as you are saying that a number is the same as an assertion, but the meaning is clear enough. Is there a more elegant way to phrase that as tersely?
f(2020)-f(2017) must be divisible by 2020-2017=3. Since f(2017)=1, f(2020)-1 must be divisible by 3. That excludes all the answers except 2017.
2016=672*3
2017-1 is not divisible by 3
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2017-1=2016, which is definitely divisible by 3.
F(x) = 1 + A(x)(x-2017) Where A(x) is also an integer polynomial. Thus f(2020) = 1 + A(2020)*3 Thus it is one more than a multiple of 3. The only option is 2017 = 1 + 3(672)
Simplest and cleanest solution! Thanks for sharing!
I liked it the most
First note that 2 0 2 0 = 2 0 1 7 + 3 . Since the polynomial has integer coefficients, f ( 2 0 2 0 ) m o d 3 = f ( 2 0 1 7 ) m o d 3 = 1 . Only 1 9 9 9 = 1 m o d 3 . Therefore the answer is 1 9 9 9
Nice solution. If only if it was right.
Figured the simplest polynomial would be a linear one, so an increment of 3 must lead to "3 times some number + 1".
Therefore, if you divide the numbers by 3, you must get 1 as the remainder.
Only 2017 does that.
f(2020) - 1 must be divisible by 3 therefore only 2017 = 3x672 + 1 can be a possible value among those listed as choices (2015,2016,2017,2018).
Proof:
If f(2017) = 1 then there is a polynomial g(x) such that f(x) = g(x) + 1 when g(x) still has integer coefficients.
g(x) has a root at 2017 and can be factored to (x-2017)h(x) for some polynomial h(x).
At x = 2020, x-2017 = 3. Therefore g(2020) is divisible by 3 and f(2020) has the form 3N + 1 for some integer N.
Since f(2020) must be in the form 3N + 1 only 2017 = 3 x 672 + 1 qualifies as a possible value of f(2020) among the choices given.
Since f has integer coefficients, for any integers a , b , we must have a − b ∣ f ( a ) − f ( b ) . To prove this, let f ( x ) = c n x n + c n − 1 x n − 1 + . . . + c 1 x + c 0 , where all the c i s are integers. We have
f ( a ) − f ( b ) = ( c n a n + c n − 1 a n − 1 + . . . + c 1 a + c 0 ) − ( c n b n + c n − 1 b n − 1 + . . . + c 1 b + c 0 ) = c n ( a n − b n ) + c n − 1 ( a n − 1 − b n − 1 ) + . . . + c 1 ( a − b ) = ( a − b ) ( z ) ,
where z is some expression that evaluates to an integer. We used the fact that a − b ∣ a k − b k for all positive integers k to take out the factor of a − b . Thus, we conclude that a − b ∣ f ( a ) − f ( b ) .
Going back to the problem, we have
2 0 2 0 − 2 0 1 7 = 3 ∣ f ( 2 0 2 0 ) − f ( 2 0 1 7 ) = f ( 2 0 2 0 ) − 1 .
Among the answer choices given, only f ( 2 0 2 0 ) = 2 0 1 7 makes this statement true.
For completeness, we can find that f ( x ) = 6 7 2 ( x − 2 0 1 7 ) + 1 is a polynomial function with integer coefficients such that f ( 2 0 1 7 ) = 1 and f ( 2 0 2 0 ) = 2 0 1 7 .