+28 Let f(x) be a polynomial with integer coefficients. There exist distinct integers p,q,r,s,t such thatf(p)=f(q)=f(r)=f(s)=10
and f(t)>10What is the smallest possible value of f(t)
+805 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.
