18. In a polynomial of $5^{\text{th}}$ degree $p(x)=a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f$, each of the coefficients $a, b, c, d, e, f$ is 1 or -1. Knowing that $p(2)=11$, what is the value of $p(3)$?
(A) 178
(B) 244
(C) 126
(D) 142
(E) 196
(18) The correct answer is $(D)$.

Since $2^{5}=32$ is greater than the sum of $1,2,4,8,16$, the sign of $p(2)$ matches the sign of $a$. Knowing that $p(2)=11>0, a$ must be positive, and thus $a=1$. We can repeat the same reasoning for the polynomial $q(x)=p(x)-x^{5}=b x^{4}+c x^{3}+d x^{2}+e x+f$ : since $q(2)=11-32=-21$, it follows that $b<0$ and thus $b=-1$. Continuing, we easily obtain $p(x)=x^{5}-x^{4}-x^{3}+x^{2}-x+1$. At this point, it is straightforward to calculate $p(3)=3^{5}-3^{4}-3^{3}+3^{2}-3+1=243-81-27+9-3+1=142$. Problem proposed by Matteo Rossi.