any number mod 3 is congruent to the sum of its digits, and observe that any number $x$ Mica has written verify $x\equiv 1(3)$ then any permutation of the digits of this number belongs too to the list. So you can underlined any number and all of its permutation except the ones like this $a0..0$ with $a\in [[1,9]]$ because it is giving a number with less digits. Sorry I'm pbably missing sth, what the problem asks exactly is not totally clear for me?

BlayzyMath wrote: any number mod 3 is congruent to the sum of its digits, and observe that any number $x$ Mica has written verify $x\equiv 1(3)$ then any permutation of the digits of this number belongs too to the list. So you can underlined any number and all of its permutation except the ones like this $a0..0$ with $a\in [[1,9]]$ because it is giving a number with less digits. Sorry I'm pbably missing sth, what the problem asks exactly is not totally clear for me? You have to underline numbers that have only one digit but used repeatedly. For example, you can underline $1111$, $4444$ and $7777$.

why ? why not 1231 and 1123 they have the same digits and are in the interval

BlayzyMath wrote: why ? why not 1231 and 1123 they have the same digits and are in the interval Each number underlined by Facu has to have all its digits the same. You can't underline $1231$ because $1\neq 2$.

Notice that Mica's list is contain numbers that are $\equiv 1 \mod 3$ Bashing CRT - 2 digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 11$ You'll get $n = 22 + 33k$ $n = \{22, 55, 88\}$ - 3digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 111 \rightarrow n \equiv 0 \mod 3 \text{(contradiction)}$ No solution - 4 digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 1111$ You'll get $n = 1111 + 3333k$ $n = \{1111, 4444, 7777 \}$ - 5 digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 11111$ You'll get $n = 22222 + 33333k$ $n = \{22222, 55555, 88888 \}$ So, the underlined numbers are $\{22, 55, 88, 1111, 4444, 7777, 22222, 55555, 88888 \}$

ok thanks guys

DavyDuf wrote: Notice that Mica's list is contain numbers that are $\equiv 1 \mod 3$ Bashing CRT - 2 digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 11$ You'll get $n = 22 + 33k$ $n = \{22, 55, 88\}$ - 3digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 111 \rightarrow n \equiv 0 \mod 3 \text{(contradiction)}$ No solution - 4 digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 1111$ You'll get $n = 1111 + 3333k$ $n = \{1111, 4444, 7777 \}$ - 5 digits --> $n \equiv 1 \mod 3 \ \text{ and } n \equiv 0 \mod 11111$ You'll get $n = 22222 + 33333k$ $n = \{22222, 55555, 88888 \}$ So, the underlined numbers are $\{22, 55, 88, 1111, 4444, 7777, 22222, 55555, 88888 \}$ That's correct, but a bit of an overkill. If $n$ has $2$ digits. Let them be equal to $a$, then $2a\equiv 1\pmod 3$, so $a\equiv 2\pmod 3$. If $n$ has $3$ digits. Let them be equal to $a$, then $3a\equiv 0\pmod 3$, so no possibilities here. If $n$ has $4$ digits. Let them be equal to $a$, then $4a\equiv 1\pmod 3$, so $a\equiv 1\pmod 3$. If $n$ has $5$ digits. Let them be equal to $a$, then $5a\equiv 1\pmod 3$, so $a\equiv 2\pmod 3$. Then you get your answer.

$\color{blue} \boxed{\textbf{SOLUTION}}$ First, note that, all numbers in the sequence $a\equiv 1\pmod 3$ $\implies a=3k+1$ $\color{black} \boxed{\textbf{Case 1}}$ Let, $n=11m$ So, we want, $11m=3k+1$ $11m\equiv 2m\pmod 3$ $\implies m=2,5,8$ $n={22,55,88}$ $\color{black} \boxed{\textbf{Case 2}}$ Let, $n=111m\equiv 0\pmod 3$ So, No Solution $\color{black} \boxed{\textbf{Case 3}}$ Let, $n=1111m\equiv m\pmod 3$ $\implies m=1,4,7$ $n={1111,4444,7777}$ $\color{black} \boxed{\textbf{Case 4}}$ Let, $n=11111m\equiv 2m\pmod 3$ $\implies m=22222,55555,88888$ $n={22222,55555,88888}$ So, $\color{red} \boxed{\textbf{n={22,55,88,1111,4444,7777,22222,55555,88888}}}$