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$$