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A polynomial is denoted by $f(x)$. The coefficients of the polynomial are positive integers. $f(1) =17$ $f(20)=421350$ Could you tell if such a polynomial is possible? If ye, find the degree of the polynomial and also it's coefficients.

My inference:Using $f(20)=421350$ we can determine, that the degree of the polynomial cannot exceed $4$. Also since every coefficient is positive, therefore each individual coefficient $< 17$. The constant coefficient independent of $x$ is $10$. Since $421350$-constant term should be divisible by $20$.

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Note that for any polynomial $P$ with integer coefficients, and any integers $a$ and $b$, $P(a)-P(b)$ is divisible by $a-b$. This is because $a^k-b^k$ is divisible by $a-b$.

Thus if there is a polynomial $f(x)$ with the given values, then $421350-17$ must be divisible by $19$. But it isn't. So no such polynomial exists, even without the positivity restriction.

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    I do not see the connection. What I wrote is that **if** $f(20)=421350$ and $f(1)=17$, where $f$ is a polynomial with integer coefficients, **then** $20-1$ must divide $421350-17$. It doesn't, which means there is no such polynomial $f(x)$.2012-11-15
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Brute force. Take your constant coefficient 10, and look at $\frac{f(20)-10}{20}=21 067$ This gives the coefficient of $x$ as 7. Total of coefficients so far is 17, which is given by $f(1)$, but you need higher terms to get $f(20)$ right.

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    Strongly agree. I upvoted, of course, but that only partly negates the effect of the incorrect downvote.2012-11-15