the answer is 2 step 1: pick k=2 step 2: pick k=P(2)+1 then read the result in base k

DottedCaculator wrote: the answer is 2 step 1: pick k=2 step 2: pick k=P(2)+1 then read the result in base k woah how does that work?!

RedFireTruck wrote: DottedCaculator wrote: the answer is 2 step 1: pick k=2 step 2: pick k=P(2)+1 then read the result in base k woah how does that work?! To get the coefficients in base $k$ representation, we need to ensure that all the coefficients are less than $k$ But say $a_i$ is the coefficient of $x^i$ in $P(x)$. Then $k =( a_0 +1) +\sum_{k=1}^{n}2^ka_k$ Hence definitely $a_i < k$

DottedCaculator wrote: the answer is 2 step 1: pick k=2 step 2: pick k=P(2)+1 then read the result in base k Sorry, I don't understand what you mean and why that's true. Take $P(x) = 9x+5$. Step 1: $P(2) = 23$. Step 2: $P(24) = 221$. What do I do from here?

@above $P(24)=221=9*24+15\implies P(x)=9x+5$

Was confused for a minute until I saw that the coefficients were nonnegative. We can just read the polynomial $a_1x^n+a_2x^{n-1}+\cdots+a_nx+a_{n+1}$ as $a_1a_2a_3a_4\cdots a_{n+1}$ in base $n$ Smart!

natmath wrote: @above $P(24)=221=9*24+15\implies P(x)=9x+5$ But $P$ is not necessarily linear

@above The example that #5 gave had $P(x)$ as linear... Of course, for a different example it could be of different degree, but I was assuming #5 wanted an explanation of the method for that particular example. Just for my own understanding, does this work: pick $k=1$ pick $k=P(1)+1$ Read the result in base $P(1)+1$

natmath wrote: Just for my own understanding, does this work: pick $k=1$ pick $k=P(1)+1$ Read the result in base $P(1)+1$ It does. For positive $k$, $P(k)+1 \ge P(1)+1 \ge 2$ and so the base is always greater than 1 and is valid. (This also answers why the need for $+1$) My question is, is it possible to do it in 1 step with some nonpositive $k$?

Dotted pr0. Few Questions: why pick k =2. Why not 7 or perhaps 1000. why pick k = P(2) +1. why read the result in base k. where do learn these things?

Wait am I doing something wrong? Let P(x) = 2x - 3. P(2) = 1. P(P(2) + 1) = P(1 + 1) = P(2) = 1. Reading 1 in base 2 we get 1. But not clearly working with what Dotted said. @below Thanks so much.

@above note that the question restricts $P(x)$ to nonnegative coefficients. Dotted's method does not work with nonnegative coefficients as you have pointed out.

coolmath_2018 wrote: Dotted pr0. Few Questions: why pick k =2. Why not 7 or perhaps 1000. why pick k = P(2) +1. why read the result in base k. where do learn these things? All those values of $k$ should work. The point is that you want an upper bound on the coefficients; no matter what $k\ge 1$ you pick, $P(k)$ will be such an upper bound. Then we know for sure that all coefficients are strictly less than $P(k)+1$, allowing us to plug that in and read the result in base $P(k)+1$.