a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5$

Let $P(x)$ be a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5.$ Determine the value of $P(6)$.

Let $P(x) = ax^4 + bx^3 + cx^2 + dx + e$. For $n=1,2,3,4,5$ I have plugged it into this polynomial and got the following — $$P(1) = a+b+c+d+e = 120$$ $$P(2) = 16a + 8b + 4c + 2d + e = 60$$ $$P(3) = 81a + 27b + 9c + 3d + e = 40$$ $$...$$ And what the problem asks for is $$P(6) = 1296a + 216b + 36c + 6d + e .$$ However, I'm not sure if all this is helping me very much. So noticing that $2P(2) = P(1) = 3P(3)$ (which is also equal to $4P(4), 5P(5)...$) From solving simultaneous equations I got that $31a + 15b + 7c + 3d + e=0$ and similarly $211a + 65b + 19c + 5d + e=0$, but they seem rather useless at this point.

So noticing that $2P(2) = P(1) = 3P(3)$ (which is also equal to $4P(4), 5P(5)...$)

You are on the right track. $$x P(x) - 120$$ is a polynomial of degree (at most) 5, and has zeros at $x= 1, 2, 3, 4, 5$, therefore $$ x P(x) - 120 = c(x-1)\cdots (x-5) $$ for some constant $c$, which can be determined by substituting $x = 0$.

Hint:

$$P(n)=n(n-1)(n-2)(n-3)(n-4)\sum_{r=0}^4\dfrac{a_r}{n-r}$$

Hint: We can write $P(x)$ in the form of $$P(x) =c_1(x-2)(x-3)(x-4)(x-5)\ +\ c_2(x-1)(x-3)(x-4)(x-5)\ +\dots+\ c_5(x-1)(x-2)(x-3)(x-4)$$