2 and 5. The equality $2008=1111+444+222+99+77+55$ is an example of a decomposition of the number 2008 as a sum of distinct numbers with more than one digit, whose representation (in the decimal system) uses only one digit.

i) Find a decomposition of this type for the number 2009.

ii) Determine for the number 2009 all possible decompositions of this type that use the smallest number of summands (the order of the summands does not matter).
Solution:

Grouping the numbers with the same number of digits, we have the equation $2009=1111 a+111 b+11 c$

where $a, b$, and $c$ are integers less than or equal to $1+2+3+\cdots+9=45$, since the addends in the decomposition must be different.

We then have $2009=182 \cdot 11+7=11(101 a+10 b+c)+b$. Therefore, $182=101 a+10 b+c+(b-7) / 11$. Since $a, b$, and $c$ are integers, it follows that $b$ must be of the form $c=11 k+7$, for some natural value of $k$ between 0 and 3 (remember that $c \leq 45$).

Substituting, we get $182=101 a+110 k+70+c+k$, from which $112=101 a+111 k+c$. The possible solutions are $(a=1, k=0, c=11)$ and $(a=0, k=1, c=1)$. In the first case, we have $b=7$, and in the second $b=18$, so one decomposition can be $2009=1111+777+66+55$, where $c=11$ has been decomposed as $6+5$.

Analyzing the solutions, we see that it is not possible to obtain a decomposition for 2009 with fewer than 4 addends, with the first type being $a=1, b=7, c=9+2=8+3=7+4=6+5$, and the second type being $a=1, b=9+8+1=9+7+2=9+6+3=9+5+4=8+7+3=8+6+4=7+6+5, c=1$.