[Agamdeep Singh]

let (c,d) be the interval in which [p(n)] = k. (k is some constant in the range of p(x))

let r lie between in the interval (c,d).

we have,

⌊P⌊P(r)⌋⌋ + r = 4⌊P(r)⌋

⌊P(k)⌋ + r = 4k                  - [eq 1]

let e be such that r+e also lies in (c,d)

⌊P⌊P(r+e)⌋⌋ + r+e = 4⌊P(r+e)⌋

⌊P(k)⌋ + r + e = 4k            - [eq 2]

[eq 2] - [eq 1]

e=0

therefore no such interval in which ⌊P(x)⌋ is constant.

 

since a polynomial is continuous, every polynomial has to take values from some integer m to m + 1 and in the  interval [ m , m+1) , ⌊P(x)⌋ = constant.  [for p(x) in that range.]

but since there is no such interval in which ⌊P(x)⌋ is constant, there is no such polynomial.

 

[Agamdeep Singh]

Thanks sir

[Agamdeep Singh]

On the website, the answer is 401

[Agamdeep Singh]

Here's the full solution I did:

Finding the function:

finding the range:

original question (AIME 1997 P12) asked to find the unique number not in the range of the function which we found as 58 way early in the ratio -b/c

[Agamdeep Singh]

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