1. Isaac writes down a three-digit number. None of the digits is a zero. Isaac gives his sheet with the number to Dilara, and she writes under Isaac's number all the three-digit numbers that can be obtained by rearranging the digits of Isaac's number. Then she adds up all the numbers on the sheet. The result is 1221. What is the largest number that Isaac could have written down?
B1. 911 We are going to look at the possible starting numbers of Isaac. We distinguish three cases: there can be one, two, or three different digits in Isaac's starting number.

- If only one digit occurs in the starting number, then the starting number is of the form $\overline{a a a}$ (by which we mean the number consisting of the digit $a$ three times). Dilara then writes down no numbers, because there is no other number you can get by writing the digits of $\overline{a a a}$ in a different order. The sum is then $\overline{a a a}$ and that is always less than 1221.
- If two different digits occur in the starting number, then the starting number is of the form $\overline{a a b}, \overline{a b a}$, or $\overline{b a a}$ (with $a \neq b$). If Isaac's number is one of these three numbers, then Dilara writes down the other two. The sum then becomes $\overline{a a b}+\overline{a b a}+\overline{b a a}$. We see that it must end in a 1, so $b+a+a=11$ or $b+a+a=21$. But if $b+a+a=21$, then we see that the total sum is already at least 2100 (because the sum of the first digits is also 21) and that is more than 1221. The only option is $b+a+a=11$.

To find the largest possible starting number, we want to start with the largest possible digit. We see that $a$ can be at most 5, but $b$ can be 9 if $a$ is 1. We try $b=9$ and $a=1$ and we look at the starting number $\overline{b a a}=911$. Then Dilara writes down the numbers 191 and 119, and then we see that the sum $911+191+119$ is indeed equal to 1221.

- If three different digits occur in the starting number, then the starting number is of the form $\overline{a b c}$ (with $a, b$, and $c$ different). Dilara then writes down the numbers $\overline{a c b}, \overline{b a c}, \overline{b c a}, \overline{c a b}$, and $\overline{c b a}$. If we now look at these six numbers, we see that for the last digit, the digit $a$ appears twice, the digit $b$ appears twice, and the digit $c$ appears twice. The last digit of the sum is thus the last digit of $2 a+2 b+2 c$. This digit is always even, while the last digit of 1221 is odd. The sum can therefore never be equal to 1221.

We have seen that Isaac's starting number must consist of two different digits and that the largest starting number for which the sum is 1221 is the number 911.