5. On a table, there are a number of cards. Each card has a number written on it. The operation "pick and replace" involves the following: two random cards are picked from the table and replaced by one new card. If the numbers on the two picked cards are $a$ and $b$, then the number on the new card is $a + b + ab$.

If we start with ten cards on which the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are written respectively, what value(s) can the number have that remains on the only card left on the table after nine "pick and replace" operations? Prove your answer.
5. Consider the number you get by increasing the numbers on the cards lying on the table by 1 and then multiplying them all together. Call this large number the VALUE of the cards on the table. At the beginning, the VALUE is equal to $(1+1)(2+1)(3+1) \ldots(10+1)=2 \times 3 \times 4 \times \ldots \times 11=11$ ! and at the end, the VALUE is equal to the number on the last card plus 1.

If we take two cards with the numbers $a$ and $b$ and replace them with one card with the number $a+b+a b$, then the VALUE of the cards on the table does not change, because $(a+1)(b+1)=(a+b+a b)+1$. This means that the VALUE at the beginning and at the end are the same: the number on the last card is therefore $11 !-1=39916800-1=39916799$.

Copyright Dutch Mathematical Olympiad 2003