Let f(x) be a polynomial with integer coefficients such that f(5) = f(7) = 20. What is the smallest possible value of |f(0)|?
A bit of a trick question! If f(x) is the constant polynomial 20, then you get |f(0)| = 20, which is the minimum. 
+118608 If it did not have integer coefficients the answer would be 0.
BUT
with integer coefficients I do not know ???
If I had time and inclination I would go watch some videos on Rational Root Theorum and maybe I could figure it out.
BUT
I'd really like someone to show my how to do it here. 
Any takers?
The minimum is, I think, 15.
Suppose that \(f(x)=a+bx+cx^{2}+...\)
then \(|f(0)|=a.\)
Substituting x=5 and x=7 gets you
\(20-a=5b+25c+...\)
and
\(20-a=7b+49c+...\)
For integer a, b and c, it's necessary that 20 - a should be divisible by both 5 and by 7.
The smallest, in magnitude, a, is -15.
The quadratic \(f(x)= -15+12x-x^{2}\)
works, I haven't looked for a cubic.
+26367 Let f(x) be a polynomial with integer coefficients such that \(f(5) = f(7) = 20\).
What is the smallest possible value of \(|f(0)|\)?
a)
\(f(x) = 20 + (5-x)(7-x) \qquad \Big(~ f(x)=x^2-12x+55 ~\Big)\)
\(\begin{array}{|rcll|} \hline f(0) &=& 20 + (5-0)(7-0) \\ f(0) &=& 20 + 35 \\ f(0) &=& 55 \\ \mathbf{|f(0)|} &=& \mathbf{55} \\ \hline \end{array}\)
b)
\(f(x) = 20 - (5-x)(7-x) \qquad \Big(~ f(x)=-x^2+12x-15 ~\Big)\)
\(\begin{array}{|rcll|} \hline f(0) &=& 20 - (5-0)(7-0) \\ f(0) &=& 20 - 35 \\ f(0) &=& -15 \\ \mathbf{|f(0)|} &=& \mathbf{15} \\ \hline \end{array}\)
The smallest possible value of \(|f(0)|\) is 15
