## Task 34/80

Determine $s_{40}$ of an arithmetic series of 1st order with:

1. $a_{1}=10 a+b=x$ with $a ; b \in N ; 1 \leq a \leq 9 ; 0 \leq b \leq 9$,
2. $a_{40}=10 c+d=y$ with $c ; d \in N ; 0 \leq c ; d \leq 9$,
3. $s_{40}=\sum_{i=1}^{40} a_{i}=1000 a+100 b+10 c+d$
It is $s_{n}=\left(a_{1}+a_{n}\right) \cdot \frac{n}{2}$, thus

$$
s_{41}=(10 a+b+10 c+d) \cdot 20=1000 a+100 b+10 c+d
$$

This leads to the Diophantine equation $800 a+80 b-190 c-19 d=0$, which can also be written in the form

$$
80(10 a+b)=19(10 c+d) \quad ; \quad 80 x=19 y
$$

$(x ; y \in N, 10 \leq x ; y \leq 100)$. It follows that $y=4 x+\frac{4 x}{19}$, so $x \equiv 0$ (mod 19). Let $x=19 k$ with $k \in N, k \geq 1$. Then $y=4 \cdot 19 k+4 k=80 k$. Only $k=1$ yields a $y$ within the defined range. Therefore,

$$
x=10 a+b=19 ; \quad y=10 c+d=80 \quad ; \quad a=1, b=9, c=8, d=0
$$

and consequently $s_{40}=1980$ is the only solution.