
4. The equality

$$
2008=1111+666+99+88+44
$$

is a decomposition of the number 2008 into the sum of several pairwise distinct multi-digit numbers, each of which is represented (in the decimal system) using the same digits. Find

a) at least one such decomposition of the number 8002,

b) all such decompositions of the number 8002 which have the least possible number of summands (the order of summands is irrelevant).


Solution. a) An example of such decomposition is

$8002=3333+999+888+777+666+555+333+99+88+77+66+55+44+22$.

In the second part we show that this is the only such decomposition of the number 2008 into 14 summands, and that no decomposition into a smaller number of summands exists.

b) A number of the form $\overline{a a a a}$, or $\overline{a a a}$, or $\overline{a a}$, is the $a$-multiple of the number 1111, 111, or 11, respectively. Hence any decomposition of the number 8002 of the type considered can be upon summing the summands with the same number of digits brought into the form

$$
8002=1111 k+111 l+11 m
$$

where $k, l, m$ are nonnegative integers not exceeding $1+2+\cdots+9=45$ (since the summands that have been summed were pairwise distinct).

We rewrite the last equality as

$$
\begin{gathered}
8002=727 \cdot 11+5=11(101 k+10 l+m)+l, \\
727=101 k+10 l+m+\frac{l-5}{11} .
\end{gathered}
$$

This implies (in view of $l \leqslant 45$ ) that $l=11 q+5$ where $q \in\{0,1,2,3\}$. We thus obtain the equality

$$
677=101 \cdot 6+71=101(k+q)+10 q+m
$$

from which it clearly follows that $k+q=6$ and $10 q+m=71$. The last system has, under the above restriction on $q$, the only solution $q=3, k=3, m=41$. For $l$ this gives $l=38$.

To obtain the desired decomposition it remains to decompose these numbers $k, l$, $m$ into sums of one or more distinct single-digit summands. Since we have exactly the nine summands $9+8+7+6+5+4+3+2+1$ to choose from, whose sum equals 45 , it is evidently simpler to list the decompositions for $k, 45-l$ and $45-m$ (the decompositions of $l$ and $m$ can then be obtained upon omitting the used summands from the sum $1+2+\cdots+9)$ :

$$
\begin{gathered}
k=3=1+2, \\
45-l=7=1+6=1+2+4=2+5=3+4, \\
45-m=4=1+3 .
\end{gathered}
$$

We have thus found all the $2 \cdot 5 \cdot 2$ possible decompositions of the number 8002 , each of which has at least $1+6+7=14$ summands, the only decomposition into 14 summands being the one given in part a) of this solution.
