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本文想通过对若干竞赛试题的分析,讲一些解题方法。下面分几个方面讨论,限于篇幅这里将不讨论竞赛中大量出现的几何题。一有关整数性质的题这类题目在竞赛中极多,它们涉及到数的整除性:带余表示(设a,b为任意整数,b>0。则有唯一的整数m与r,使得a=mb+r,0≤r<6);质数:数的奇偶性等等。例1 一个六位数,如果它的前半部分三位数字与后半部分三位数字完全相同,顺序也相同。则7、11、13必是此六位数的约数。做题首先是审题。依题意所设六位数应是 (?) 由于7、11、13都是质数。且7·11·13=1001,所以本题无非是要证明N被1001整除,为此,只要注意到 (?)即证得本题。
This article wants to explain some problem solving methods through the analysis of several contest questions. The following discussion in several aspects, due to limited space will not discuss the geometric problems that appear in the contest. A question about the nature of integers, such topics in the competition are very many, they involve the divisibility of the number: with the remainder of the expression (set a, b for any integer, b> 0. There is a unique integer m and r, so that a =mb + r, 0 ≤ r <6); prime number: the parity of the number and so on. Example 1 A six-digit number if its first three digits are exactly the same as the last three digits and the sequence is the same. Then 7, 11, 13 must be the six-digit divisor. The first thing to do is to examine questions. The six-digit number should be (?) because the 7, 11 and 13 are all prime numbers. And 7 · 11 · 13 = 1001, so this question is nothing more than to prove that N is divisible by 1001, for which, as long as you notice (?) That is the proof of this question.