Find the integer n

Question

Let $a$ and $b$ be two integers such that $10a+b=5$ and $p(x)=x^2+ax+b$. Find the integer $n$ such that $p(10)p(11)=p(n).$ Please tell how to proceed.

Answer 1

Well, $$p(10)=100+10a+b=100+5=105,$$ and $$p(11)=121+11a+b=121+a+10a+b=121+a+5=126+a.$$ Hence, $$\begin{align}p(10)p(11) &= 13230+105a\\ &= 13225+105a + 5\\ &= 13225+105a+10a+b\\ &= 13225+115a+b\\ &= p(115).\end{align}$$

Answer 2

You have $p(x) = x^2+ax+b$ and you're trying to solve for $n$ such that $p(10)p(11) = p(n)$.

Substituting the formula for $p(x)$ into the equation, we get $(100+10a+b)(121+11a+b) = n^2+na+b$.

Using the fact that $10a + b = 5$, we get $(100+5)(121+a+5) = n^2+(n-10)a+5$, or $13230+105a = n^2+(n-10)a+5$.

We suspect that $n = 115$, so we check if $n^2 + 5 = 13230$, and indeed it is.

Answer 3

$$p(x) = x^2+ax+5-10a$$ $$p(10) = 100+10a+5-10a = 105$$ $$p(11) = 121+a+5 = 126+a$$ $$p(10)p(11) = 105*126+105a$$ $$p(n) = n^2+a(n-10)+5 = 105(126+a)$$ $$n = 105+10=115$$