I attacked the problem by first setting the obvious as P(x) = c1xn+c2xn-1+.....+cn with the c's being the constants.

P(1) = 5 tells me that the sum of all the coefficients of the polynomial is 5, and immediately a matter to observe would be n < 4 because the next part, P(P(1)) = 177 means P(5) = 177 and 54 > 177 for obvious reasons.

So P simplifies to P(x) = x3+c1x²+c2x+c3. It's trivial to see x3 won't have a coefficient. My next steps involved basically trial and error-ing as the sum of coefficients = 5 and using every different case where I change the degree, remove some terms and overall experiment with the coefficients and x terms.

In the end, I found P(x) = x3+2x²+2 to be the only polynomial satisfying the given conditions and the answer to the question is P(10) = 103+2(10)²+2 = 1202.

My problem is I used a trial and error approach to the problem and found my solution, which I'm unsure is the only solution. Is there a more elegant approach to the problem and is my solution correct? Thanks~

alphayash Jul 17, 2022

#1**0 **

The problem got cut off in your title, could you please show us the complete question?

Voldemort Jul 17, 2022

#2**+1 **

The question has to be in the question section.

A suitable title might have been

Polynomials, sum of roots (I assume you want to find the sume of all possible roots..)

It is really good that you have tried to show what you did :)

Melody Jul 17, 2022

#3**+1 **

[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?

this was the whole question

alphayash Jul 17, 2022

#5**+2 **

__Question:__

__[Polynomials] P(x) is a polynomial in x with non-negative integer coefficients. If P(1) = 5 and P(P(1)) = 177, what is the sum of all possible values of P(10)...?__

To begin with, your solution is correct, and is actually a quick way to solve this problem. That is, It is efficient.

However, I think what you are trying to say, is there a way to write the solution "more rigorously", and the answer to this is: yes.

Let \(P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0\), where \(a_n,a_{n-1},...,a_0 \ge 0\) (Given.)

Since: \(P(1) = 5 \iff a_n+a_{n-1}+...+a_0=5\), which does not tell us that much.

Next, \(P(P(1)) = P(5) =a_n*5^n+a_{n-1}*5^{n-1}+...+a_0=177\)

Notice: \(5^3=125,5^4=625\); hence, \(P(x)\) is at most a cubic polynomial.

Now,

Let: \(P(x)=ax^3+bx^2+cx+d\) , where \(0\le a,b,c,d \le 5\)

We know: \(a+b+c+d=5\) (1)

And, \(125a+25b+5c+d=177\) (2)

Notice from (2):

\(125a \le 177 \implies a \le 1.416 \implies a=1\text{ or, } a=0\)

Next, consider the __ case a=1__ :

(2) and (1) becomes:

\(25b+5c+d=52 \text{, and } b+c+d=4\)

Using the same idea as before:

\(25b \le 52 \implies b \le 2.09 \implies b=2,1,0\)

But, \(b=1,0\) is rejected. Because observe that if b=0, then 5c+d = 52 (and this is only true for large c and d, which we are not given.)

Proof: \(c,d\le 5 \implies 5c \le 25, d\le 5 \implies 5c+d \le 25+5=30\), so it is at most 30 (Which, by the way, is even unattainable!).

Thus, \(b=2\) only.

Next, we substitute b=2 in (1) and (2) to get:

\(c+d=2\text{ , }5c+d=2 \implies 5c\le 2 \implies c=0,d=2\)

So: \(a=1,b=2,c=0,d=2\) is a valid solution.

Thus, \(P(x)=x^3+2x^2+2 \implies P(10)=1000+200+2=1202\), as you found.

Next, consider the __ case a=1__ :

\(25b+5c+d=177 \implies b \le 7\), we see that, this is impossible. As, b is at most 5, giving 125, but c and d are forced to be 0. Hence, this case is rejected.

Proof:

\(25b \le 125, 5c \le 25, d \le 5 \implies 25b+5c+d \le 155\), which again, is not attainable anyway.

Therefore,\(\text{ }P(x)=x^3+2x^2+2\) is the only possible polynomial with the given conditions, and 1202 is the only valid solution.

I hope this helps!

Guest Jul 17, 2022

#7**+1 **

Ooh! I just saw it now. It is really elegent! But I think my approach is more general; as writing 177=125+50+2 depended on "177" probably If it was other number, then there might be many ways to write a number "A" as addition of other numbers, so more cases to try.

But overall, that was fantastic observation!!!!!!

Guest Jul 17, 2022

#4**+1 **

If P(x) is a polynomial with all positive integer polynomials AND the constant is also positive (which may not be intended)

and

P(1)=5 then all the coeficients and the constant must add to 5.

P(5)=177 = 125+50+2 = 125+ 2*25 + 2 so the polynomial is

\(p(x)=x^3+2x^2+2\)

That is the only possibility. (assuming that the constant is also a positive integer)

Melody Jul 17, 2022