
00.1. In how many ways can the number 2000 be written as a sum of three positive, not necessarily different integers? (Sums like $1+2+3$ and $3+1+2$ etc. are the same.)


Solution. Since 3 is not a factor of 2000 , there has to be at least two different numbers among any three summing up to 2000 . Denote by $x$ the number of such sums with three different summands and by $y$ the number of sums with two different summands. Consider 3999 boxes consequtively numbered fron 1 to 3999 such that all boxes labelled by an odd number contain a red ball. Every way to put two blue balls in the even-numbered boxes produces a partition of 2000 in three summands. There are $\binom{1999}{2}=999 \cdot 1999$ ways to place the blue balls. But htere are $3!=6$ different placements, which produce the same partition of 2000 into three different summands, and $\frac{3!}{2}=3$ different placements, which produce the same partition of 2000 into summands two which are equal. Thus $6 x+3 y=$ 1999.999. But $y=999$, because the number appering twice in the partition can be any of the numbers $1,2, \ldots, 999$. This leads to $x=998 \cdot 333$, so $x+y=1001 \cdot 333=333333$.
