## Task 17/73

$$
f(x)=\sum_{k=0}^{5} a_{k} x^{k}
$$

is a polynomial of degree 5. Determine integer coefficients $a_{k}>0$ such that at most one of them is divisible by 120, but $f(x)$ is divisible by 120 for all integer $x>=1$.
It is $120=2 \cdot 3 \cdot 4 \cdot 5$. The polynomial is certainly divisible by 120 for all integers $x \geq 1$ if it can be represented as a product of 5 consecutive positive integers. Among 5 consecutive positive integers, there is always at least one divisible by 2, at least one divisible by 3, at least one divisible by 4, and exactly one divisible by 5. Therefore, a representation of the form

$$
f_{1}(x)=x(x+1)(x+2)(x+3)(x+4)+a_{0} \quad \text { or } \quad f_{2}(x)=(x+1)(x+2)(x+3)(x+4)(x+5)
$$

suggests itself. We obtain

$$
f_{1}(x)=x^{5}+10 x^{4}+35 x^{3}+50 x^{2}+24 x+120
$$

$\left(a_{0}=120\right.$ results from the consideration that $a_{0}$ must be divisible by 120),

$$
f_{2}(x)=x^{5}+15 x^{4}+85 x^{3}+225 x^{2}+274 x+120
$$

A general approach is obtained with

$$
f_{n}(x)=(x+n+1)(x+n+2)(x+n+3)(x+n+4)(x+n+5)
$$

where $n$ is a non-negative integer.