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《中学数学》1983年第4期问题征解中有这样一题,求证 1+2+3+…+1983|1~5+2~5+3~5+…+1983~5。事实上,我们有一般的结论:1°。1+2+3+…+n|1~5+2~5+3~5+…+n~5,甚至更一般的结论:2°。1+2+3+…+n|1~(2k+1)+2~(2k+1)+3~(2k+1)+…+n~(2k+1)。这里n、k为任意自然数,为了证明这一结论,我们要用到整数的两个性质。性质1。两个连续整数必互质。性质2。如果(p,q)=1,p|m,q|m则pq|m、((p,q)表示p与q的最大公约数)。此二性质都很容易用反证法证明,这里从略。我们来证明上述结论2°。证∵ 1+2+…+n=(1/2)n(n+1),记 S_(2k+1)(n)=1~(2k+1)+2~(2k+1)+…+n~(2k+1),
There is such a question in the problem solving of the “Middle School Mathematics” 1983 No. 4 issue. Proof 1+2+3+...+1983|1~5+2~5+3~5+...+1983~5. In fact, we have the general conclusion: 1°. 1+2+3+...+n|1~5+2~5+3~5+...+n~5, even more general conclusion: 2°. 1+2+3+...+n|1~(2k+1)+2~(2k+1)+3~(2k+1)+...+n~(2k+1). Here, n and k are arbitrary natural numbers. In order to prove this conclusion, we need to use two properties of the integer. Nature 1. Two consecutive integers must be relatively prime. Nature 2. If (p,q)=1,p|m,q|m then pq|m, ((p,q) denotes the greatest common divisor of p and q). Both of these properties are easily proved by counter-evidence. This is abbreviated here. Let’s prove the above conclusion 2°. ∵ 1+2+...+n=(1/2)n(n+1), note S_(2k+1)(n)=1~(2k+1)+2~(2k+1)+... +n~(2k+1),