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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Permutation and Combination

Olympiad workbook · IOQM · Permutation and Combination

Consider the set F of all polynomials whose coefficients are in the...

Consider the set FF of all polynomials whose coefficients are in the set of {0,1}\{0,1\}. Let q(x)=x3+x+1q(x)=x^3+x+1. The number of polynomials p(x)p(x) in FF of degree 14 such that the product p(x)q(x)p(x) q(x) is also in FF is:

Correct answer

50

Step-by-step solution

p(x)q(x)=(x14+…)(x3+x+1)p(x)=x14→1 case p(x)=x14+x2⇒α=10,9,8,…,10→11 case p(x)=x14+xα+xβ\begin{aligned} & p(x) q(x)=\left(x^{14}+\ldots\right)\left(x^3+x+1\right) \\ & p(x)=x^{14} \rightarrow 1 \text { case } \\ & p(x)=x^{14}+x^2 \\ & \Rightarrow \alpha=10,9,8, \ldots, 10 \rightarrow 11 \text { case } \\ & p(x)=x^{14}+x^\alpha+x^\beta\end{aligned} \left.α=10,β=6,5,4,3,2,1,0α=9,β=5,4,3,2,1,0α=8,β=4,3,2,1,0α=7,β=3,2,1,0α=6,β=2,1,0α=5,β=1,0α=4,β=0\begin{array}{cc}\alpha=10, & \beta=6,5,4,3,2,1,0 \\ \alpha=9, & \beta=5,4,3,2,1,0 \\ \alpha=8, & \beta=4,3,2,1,0 \\ \alpha=7, & \beta=3,2,1,0 \\ \alpha=6, & \beta=2,1,0 \\ \alpha=5, & \beta=1,0 \\ \alpha=4, & \beta=0\end{array}\right\} 25casescasesp(x)=x^{14}+x^\alpha+x^\beta+x^r\left.α=10,β=6,r=2,1,0α=10,β=5,r=1,0α=10,β=4,r=0\begin{array}{llr}\alpha=10, & \beta=6, & r=2,1,0 \\ \alpha=10, & \beta=5, & r=1,0 \\ \alpha=10, & \beta=4, & r=0\end{array}\right\} 6 cases \left.α=9,β=5,r=1,0β=4,r=0\begin{array}{ll}\alpha=9, & \beta=5, \quad r=1,0 \\ \beta=4, & r=0\end{array}\right\} 3casescases\alpha=8, \beta=4, r=0\} 1casescases Hence, total case =1+11+28+6+3+1=50 cases. \begin{aligned} & \text { Hence, total case }=1+11+28+6+3+1 =50 \text { cases. }\end{aligned}$
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