Problem 1893

First, we show by induction on the length of $n$ that if $10n$ is bright, then $n$ is bright as well. Assume that for one digit shorter numbers the statement holds. If after deleting $0$ we obtain a bright divisor, the statement holds trivially. If the bright divisor of $10n$ is obtained after deleting some other digit, then in the end of this divisor we still have $0$, i.e., it can be written as $10d$. By the induction hypothesis, $d$ is bright. But then after deleting from $n$ the corresponding digit, we get a bright divisor $d$, which means that also $n$ is bright.

Next we show that any bright divisor of at least two-digit bright number not ending with $0$ can be obtained by deleting the first or the second digit.

Let now $n$ be a 5-digit bright number not ending with $0$ and let $d$ be its bright divisor. We consider two cases depending on which digit is deleted to obtain $d$.

1) If $d$ is obtained by deleting the first digit of $n$, then $d \mid n - d = 10^4 \cdot x$, where $x$ is the deleted digit. As $d$ does not end with $0$, it is not divisible either by $2$ or $5$. If $d$ is not divisible by $5$, then $d \le 2^4 \cdot x \le 2^4 \cdot 9 < 10^3$, contradiction. Hence $d$ is odd and $d \mid 5^4 \cdot x$. Since $d$ is four-digit not ending with zero, it must divide one of the numbers $1875$, $3125$, $4375$, $5625$. We may leave out $3125$, because in this case the first digit of $n$ must be $x = 5$, but the last digit is $5$ as well. The four-digit divisors of the remaining numbers are $1125$, $1875$, $4375$, $5625$. The first and the last number contain equal digits, from the other two numbers we cannot obtain a divisor by deleting the first or the second digit.

2) If a bright divisor is obtained by deleting the second digit, then $d \mid n - d = 10^3 \cdot z$, where $z$ is at most two-digit. Since $d$ does not end with $0$, it is not divisible either by $2$ or $5$, implying that $\frac{n-d}{d}$ is divisible either by $2^3$ or by $5^3$. Since $n$ and $d$ start with the same digit $a$, we have $\frac{n}{d} < \frac{(a+1) \cdot 10^4}{a \cdot 10^3} = 10 + \frac{10}{a} \le 20$, implying that $\frac{n-d}{d}$ can be only $8$ or $16$, in both cases $5^3 \mid d$. We can write $d = 1000a + 125r$, where $r \in \{1,3,5,7\}$. If $\frac{n-d}{d} = 16$, or equivalently $n = 17d$, then $n = 17000a + 2125r = 1000(17a + 2r) + 125r$. Since $n$ starts with $a$ and $125r < 1000$, we get $17a + 2r < 10(a + 1)$ yielding $7a + 2r < 10$. This gives $a = 1, r = 1$, and $d = 1125$, which is not bright. If $\frac{n-d}{d} = 8$, or equivalently $n = 9d$, then $n = 9000a + 1125r = 1000(9a + r) + 125r$. Since $n$ starts with $a$, we get $9a + r + \frac{r}{8} > 10a$, yielding $r > \frac{8}{9}a \ge a - 1$, i.e., $r \ge a$. Leaving out numbers with repeated digits, we get $d \in \{1375, 1625, 1875, 2375, 2875, 3625, 3875, 4625, 4875, 6875\}$. Among these numbers only $1625$ is bright, deleting the first or the second digit from other candidates does not give a divisor. A check shows that $9d = 14625$ is bright as well.

Therefore $14625$ is the only 5-digit bright number not ending with $0$. Let now $n$ be arbitrary 6-digit bright number. If $n$ ends with $0$, then deleting $0$ we obtain a 5-digit bright number not containing $0$, whence $n = 146250$. If $n$ is not ending with $0$, then after deleting the first or the second digit we obtain a bright divisor $d$ not ending with $0$. Thus $d = 14625 = 117 \cdot 125$, yielding $117 \mid n - d = 10^4 \cdot z$, where $z$ is at most 2-digit. Since $117$ and $10$ are co-prime, this is not possible. Consequently, $146250$ is the only 6-digit bright number. If $n$ was a bright 7-digit number, then by deleting its some digit we would obtain $146250$. Then also $n$ should end with $0$, which means that $\frac{n}{10}$ is a bright 6-digit number not containing $0$. But there are no such numbers. Since there exist no 7-digit bright numbers, there cannot be longer bright numbers either.