6. Find all three-digit numbers n with three different non-zero digits, which are divisible by the sum of all three two-digit numbers obtained by erasing one digit in the original number.

(Jaromír Šimša)
SOLUTION. We are looking for three-digit numbers $n=\overline{a b c}=100 a+10 b+c$, which are divisible by the sum

$$
\overline{b c}+\overline{a c}+\overline{a b}=(10 b+c)+(10 a+c)+(10 a+b)=20 a+11 b+2 c,
$$

where $\{a, b, c\}$ is a three-element subset of the set $\{1,2, \ldots, 9\}$. This happens precisely when there exists a natural number $k$ such that

$$
100 a+10 b+c=k(20 a+11 b+2 c)
$$

or

$$
(100-20 k) a=(11 k-10) b+(2 k-1) c .
$$

Since the right side of the last equation is positive, with regard to the form of the left side, we must have $k \leqq 4$. We will now examine the possible values of $k$ individually.

$\triangleright k=1$. From (1) we have the equation $80 a=b+c$, which has no solution, because $80 a \geqq 80$ and $b+c \leqq 17$.

$\triangleright k=2$. From (1) we get the equation $60 a=12 b+3 c$ or $20 a=4 b+c$. Since $4 b+c \leqq 44$, we have $a \leqq 2$. Furthermore, from the equation it follows that the number $c$ must be a multiple of four, so $c \in\{4,8\}$, which in both cases $a=1, a=2 \mathrm{~s}$ we will take advantage of.

For $a=1$ we have $20=4 b+c$ with the only suitable pair of different digits $b=3, c=8$, which corresponds to $n=138$.

For $a=2$ we have $40=4 b+c$ with the only suitable pair of different digits $b=9, c=4$, which corresponds to $n=294$.

$\triangleright k=3 . \mathrm{Z}$ (1) we have the equation $40 a=23 b+5 c$. According to divisibility by five, $b$ can only be $5$. We then get $40 a=23 \cdot 5+5 c$ or $8 a=23+c$ with two suitable pairs $(a, c)$ equal to $(3,1)$ and $(4,9)$, which correspond to two sought $n$, namely $n=351$ and $n=459$.

$\triangleright k=4$. From (1) we get the equation $20 a=34 b+7 c$ or $20(a-b)=7(2 b+c)$. Since the numbers 20 and 7 are coprime and the right side is positive, $a-b$ must be a positive multiple of seven, so $a-b=7$ and $2 b+c=20$. Such a system has no solution, because according to the first equation $b \leqq 2$, and thus $2 b+c \leqq 4+9=13$.

Conclusion. The given problem is satisfied by four three-digit numbers, namely 138, 294, 351, and 459.

GUIDING AND ADDITIONAL PROBLEMS:

N1. Determine all three-digit numbers that are eleven times greater than their digit sum. $[198$, from the equation $100 a+10 b+c=11(a+b+c)$ rearranged to the form $89 a=b+10 c$ it follows $a=1$, so $b=9$ and $c=8$.]

N2. Determine all three-digit numbers that are seven times greater than the two-digit number obtained by erasing the middle digit of the given number. $[105$, from the equation $100 a+10 b+$ $+c=7(10 a+c)$ rearranged to the form $5(3 a+b)=3 c$ it follows $c=5$.]

N3. Find all four-digit numbers $\overline{a b c d}$ for which $\overline{a b c d}=20 \cdot \overline{a b}+16 \cdot \overline{c d}$. $[65-\mathrm{C}-\mathrm{S}-1]$

D1. Determine the largest three-digit number from which, after erasing any digit, we get a prime number. $[67-\mathrm{C}-\mathrm{S}-1]$

D2. Find the smallest four-digit number $\overline{a b c d}$ such that the difference $(\overline{a b})^{2}-(\overline{c d})^{2}$ is a three-digit number written with three identical digits. $[67-C-I-1]$