Let \(f(x)\) be a polynomial with integer coefficients. There exist distinct integers \(p, q,r,s,t\) such that\(f(p) = f(q) = f(r) = f(s) = 10\)
and \(f(t)>10\)What is the smallest possible value of \(f(t)\)
Let g(x)=f(x)−10. Then g(p)=g(q)=g(r)=g(s)=0, and so g is divisible by (x−p)(x−q)(x−r)(x−s). We can find this polynomial by taking the Chinese Remainder Theorem with modulus p−q on g(q), modulus p−r on g(r), and modulus p−s on g(s). This gives us $g(x)=10(x−p)(x−q)(x−r)(x−s)+100.$Hence, $f(x)=10(x−p)(x−q)(x−r)(x−s)+110.$By Vieta's formulas, the sum of the roots of x4−10x3+35x2−50x+24=0 is 10. Hence, p+q+r+s=10. By AM-GM, $(p+q+r+s)/4≥(pqrs)1/4=101/4=2,$so p+q+r+s≥8. Hence, $f(t)=10(t−p)(t−q)(t−r)(t−s)+110≥10(8)(t−8)+110=80t−660.
The minimum value of this expression is 5040, which occurs when t=11. Therefore, the answer is 5040.