What could be the two-digit number that is 6 times larger than the sum of the digits of the number that is 7 greater than it?
Let's assume there is a number $N$ that meets the requirement, and denote its digits by $c$ and $d$, i.e., $N=\overline{c d}=10 c+d$. The number $N^{\prime}$, which is greater than 7, definitely has a different unit digit from $d$; for the rest, we need to consider three possibilities:

$$
\text { 1) } \left.d+7 \leqq 9, \text { so the tens digit of } M^{\prime} \text { is also } c: 2\right) d+7 \geqq 10, d \geqq 3
$$

In this case, there is a carry to the tens place, and either $2 a$ ) $c+1 \leqq 9$, so $N^{\prime}$ is a two-digit number, or $2 b$ ) $c+1=10, N^{\prime}$ is a three-digit number.

In case 1), $N^{\prime}=10 c+(d+7)$, and the sum of its digits is $s_{1}=c+d+7$. From $N=6 s_{1}$, we get $c=10+(5 d+2) / 4$, which is greater than 10, which is impossible. Therefore, there is no solution in this case.

In case 2), the last digit of $N^{\prime}$ is $d+7-10=d-3$. In $2 a$ ), the first digit is $c+1$, and the sum of the digits is $s_{2}=c+d-2$. From $N=6 s_{2}$, we get $c=-3+5 d / 4$. Thus, $c$ is an integer if and only if $d$ is divisible by 4: either $d=4$, or $d=8$, so $c=2$, or $c=7$ and $N=24$, or 78.

In case $2 b$ ), $c=9$, and the digits of $N^{\prime}$ are: $1,0, d-3$, and their sum is $s_{3}=d-2$. From $N=6 s_{3}$, we get $d=20.4$; this is not a digit, so there is no solution in this case.

The two numbers found indeed meet the requirement: $24+7=31$ and $24=6(3+1) ; 78+7=85$ and $78=6(8+5)$.

Kultsár Szabolcs (Hajdúböszörmény, Bocskai I. g. I. o. t.)