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在新年到来之际,特作与“1989”有关的数学趣题,为读者助兴。例1 从任意给定的1989个自然数中,总可以取出若干个数(也可以是一个),其和能被1989整除。证明设1989个自然数为a_1,a_2,a_3,……,a_(1989),构造和数: a_1, a_1+a_2, a_1+a_2+a_3, ……, a_1+a_2+a_3+……+a_(1989) 这1989个和数被1989除时,如果余数各不相同,则必有一个余数为0,此和即被
On the occasion of the new year, special math questions related to “1989” were added to the reader. Example 1 From any given 1989 natural number, it is always possible to take out several numbers (which can also be one), and the sum can be divisible by 1989. The proof establishes that the 1989 natural numbers are a_1, a_2, a_3,..., a_(1989), and the sums: a_1, a_1+a_2, a_1+a_2+a_3, ..., a_1+a_2+a_3+...+a_ (1989) When the 1989 sum is divided by 1989, if the remainder is not the same, there must be a remainder of 0, and this sum is