## Task 13/69

The five smallest, non-single-digit consecutive numbers $z_{i}$ (with $\mathrm{i}=1 ; 2 ; 3 ; 4 ; 5$) are sought, for which the following holds: $z_{i}$ is divisible by $i+4$ and has the last digit $i+4$.
Let $z_{i}=10 a+i+4$, where $a$ is a natural number. According to the problem, the following relationships must hold (where $b, c, d, e, f$ are also natural numbers)

$$
10 a+5=5 b, 10 a+6=6 c, 10 a+7=7 d, 10 a+8=8 e, 10 a+9=9 f
$$

By rearranging, we get

$$
b-1=2 a, c-1=\frac{5 a}{3}, d-1=\frac{10 a}{7}, e-1=\frac{5 a}{4}, f-1=\frac{10 a}{9}
$$

From this, it follows that $a$ must be divisible by 3, 7, 4, and 9. The least common multiple of these numbers is 252. The five sought numbers are thus 2525, 2526, 2527, 2528, 2529. Checking confirms that these numbers indeed have the required properties. At the same time, it is clear from the solution that these are the smallest five such numbers that are not single-digit.