17. 5555 children, numbered 1 to 5555 , sit around a circle in order. Each child has an integer in hand. The child numbered 1 has the integer 1 ; the child numbered 12 has the integer 21; the child numbered 123 has the integer 321 and the child numbered 1234 has the integer 4321. It is known that the sum of the integers of any 2005 consecutive children is equal to 2005 . What is the integer held by the child numbered 5555 ?
(2 marks)
有 5555 名小孩, 他們編號為 1 至 5555 , 並順序圍圈而坐。每人手上均有一個整數：編號為 1 的小孩手中的整數為 1 , 編號為 12 的小孩手中的整數為 21 , 編號為 123 的小孩手中的整數為 321 , 編號為 1234 的小孩手中的整數為 $4321 \circ$ 已知任意連續 2005 位小孩手上的整數之和均為 2005 。問編號為 5555 的小孩手中的整數是甚麼？
17. -4659
17. Let $a_{n}$ denote the number held by children $n$. Since the sum of the numbers of any 2005 consecutive children is equal to 2005 , we have $a_{n}=a_{n+2005}$ for all $n$, where the index is taken modulo 5555. In particular, $a_{0}=a_{2005 k}$ for all positive integers $k$. Since $(2005,5555)=5$, there exists a positive integer $k$ such that $2005 k \equiv 5(\bmod 5555)$. Hence $a_{0}=a_{5}$, and in general $a_{n}=a_{n+5}$ for all positive integers $n$. Therefore the sum of the numbers of any 5 consecutive children is the same, and should be equal to $2005 \times \frac{5}{2005}=5$. In particular, since $\{1,12,123$, 1234, 5555 $\}$ form a complete residue system modulo 5 , we must have $a_{1}+a_{12}+a_{123}+a_{1234}+a_{5555}=5$, and hence the answer is $5-1-21-321-4321=-4659$.