Question Number 117857 by bemath last updated on 14/Oct/20
$$\mathrm{Let}\:\mathrm{n}\in\mathbb{N}\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\: \\ $$$$\mathrm{polynomials}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{coefficients} \\ $$$$\mathrm{in}\:\left\{\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\:\right\}\:\mathrm{such}\:\mathrm{that}\:\mathrm{p}\left(\mathrm{2}\right)=\:\mathrm{n}\: \\ $$
Answered by mindispower last updated on 14/Oct/20
![let p(x)=Σ_(k≤n) a_k x^k ∀j∈[1,n−1] a_j ∈A={0,1,2,3} ,a_n ∈A−{0} p(2)=n⇒2^n a_n +2^(n−1) a_(n−1) +.....+a_0 =n 2^n ≥n,∀n∈N⇒p(2)≥2^n ≥n ⇒a_n =0 aburd ⇒p(x)=a_0 =n,∀n ∃! P such p(2)=n](https://www.tinkutara.com/question/Q117893.png)
$${let}\:{p}\left({x}\right)=\underset{{k}\leqslant{n}} {\sum}{a}_{{k}} {x}^{{k}} \\ $$$$\forall{j}\in\left[\mathrm{1},{n}−\mathrm{1}\right]\:{a}_{{j}} \in{A}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}\:,{a}_{{n}} \in{A}−\left\{\mathrm{0}\right\} \\ $$$${p}\left(\mathrm{2}\right)={n}\Rightarrow\mathrm{2}^{{n}} {a}_{{n}} +\mathrm{2}^{{n}−\mathrm{1}} {a}_{{n}−\mathrm{1}} +…..+{a}_{\mathrm{0}} ={n} \\ $$$$\mathrm{2}^{{n}} \geqslant{n},\forall{n}\in\mathbb{N}\Rightarrow{p}\left(\mathrm{2}\right)\geqslant\mathrm{2}^{{n}} \geqslant{n}\:\Rightarrow{a}_{{n}} =\mathrm{0}\:{aburd} \\ $$$$\Rightarrow{p}\left({x}\right)={a}_{\mathrm{0}} ={n},\forall{n}\:\exists!\:\:{P}\:{such}\:{p}\left(\mathrm{2}\right)={n} \\ $$