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A polynomial $f(x)$ is given. All we know about it is that all its coefficients are non-negative integers, $f(1) = 6$ and $f(7) = 3438$. Hence find the remainder when $f(x)$ is divided by $(x-3)$. I have no clue on how to go about solving this.

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The polynomial would be $x^4+ 3x^3 +x+1$. The first observation $f(1)=6$ is that all the coefficients are less than $6$. So $f(7)=3438$ is the decimal equivalent of the base 7 number formed by the coefficients $1 3 0 1 1$ ($=3438_{10}$). So by remainder theorem the remainder is $f(3)= 166$.

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    OP didn't mention about the polynomial to be of fourth order – 2017-01-24
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    It can't be more than that order.This follows from the uniqueness of representation of a given number In a particular base. – 2017-01-24
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    +1: Very nice! One adjustment I would make: "no coefficient is greater than $6$." Knowing only that $f(1)=6$ still allows (for example) that $f(x)=6$ for all $x$. – 2017-01-24
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    I still don't get it... Could you possibly solve these by just using remainder theorem and/or factor theorem? – 2017-01-24
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    I think more explanation would be nice here or at least explicitly mentioning what theorems you use so the OP can look them up. I myself have never seen the uniqueness of representation in a base used in a proof like this, and I would love to see additional references on the method – 2017-01-24
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    A question of mine: does this rely on the coefficients being natural numbers? I.e. Could there be other answers if we allowed the coeffs to span the reals? – 2017-01-24
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    Even if the coefficients were to span over integers we can get 572x-566 as a valid polynomial (and so it can be seen that the polynomial would not be unique) so the remainder can vary arbitrarily. – 2017-01-24
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    For clarification, if we had $f(1)=6$ and $f(10)=3438$, would this imply that no such polynomial exists? – 2017-01-26
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    Yes you got that right there will not exist polynomial satisfying f(1)=6 and f(10)=3438 having non negative integer coefficients. – 2017-01-28