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把由1开始的自然数依次写下去,直写到198位为止、12345678901112…那么这 198位个数用9除的余数是: (A)1,(B)6,(C)7,(D)非上述答案。这是1987年全国初中数学联赛的一道试题。本文给出一种不同于常规解法的巧妙解答。首先我们证明两个定理。定理1 设数列{a_n}的每一项都是非负整数,且a_1≠0,把由a_1开始的非负整数依次写下去,直写到第n项为止即为a_1a_2…a_n,那么正整数a_1a_2…a_n除以9的余数与S_n=a_1+a_2
Write down the natural number from the beginning of 1 and write it straight to 198, 12345678901112... Then the number of 198 digits divided by 9 is: (A)1, (B)6, (C)7, (D) Not the above answer. This is a test of the National Junior High School Mathematics League in 1987. This article presents an ingenious solution that is different from the conventional solution. First we prove two theorems. Theorem 1 Let {a_n} be a non-negative integer, and a_1≠0, write down the non-negative integers starting from a_1, and write them straight to the nth term as a_1a_2...a_n, then the positive integer a_1a_2 ...a_n divided by the remainder of 9 and S_n=a_1+a_2