Arithmetic series, find the first 80 natural numbers which sum when divided by 31 has a remainder of 17 The question is as follows: 

Find the first 80 natural numbers which sum when divided by 31 has a remainder of 17
So I tried to come up with a sum formula which is like this, I came up with everything from this point on, I don't know if it's correct:

$$ \sum_{k=1}^{80}kx=1x+2x+...+80x=s$$
I came up with this I don't know if it's correct, or if I understood the question correctly. So the next step I took was:
$$a_1=1*x$$
$$a_n=80*x$$
$$n=80$$
$Y=$ divided sum
$$$$
$$$$
Then I took the sum formula that has $a_n$ in it. I divided $\frac{n}{2}(a_1+a_n)$ with 31 and added 17 to the quotient
$$Y+17=\frac{80}{74}(x+80x)$$
Forward from here I got confused.
hint

$$u_n=a+nb$$

$$\sum_{n=0}^{79}u_n=$$
$$a+(a+b)+(a+2b)+...+(a+79b)=$$
$$80a+b(1+2+...79)=$$

$$80a+b\frac{79(79+1)}{2}=$$
$$=80.a+3106.b$$
hint

$$u_n=a+nb$$

$$\sum_{n=0}^{79}u_n=$$
$$a+(a+b)+(a+2b)+...+(a+79b)=$$
$$80a+b(1+2+...79)=$$

$$80a+b\frac{79(79+1)}{2}=$$
$$=80.a+3106.b$$