2,3,12,18,72,108,216 |
number of fractions? |
我认为答案好像不止一组。能给出你的解题过程吗? |
Your conjecture seems not right. The following is an example for 4 fractions. 1= 1+(-1/2+1/2)+(-1/3+1/3)+(-1/4+1/4) =(1-1/2)+(1/2-1/3)+(1/3-1/4)+1/4 =1/2+1/6+1/12+1/4 用这个办法是否可以写出和为 1的任何多(not a product of two consecutive numbers)个分子为1而分母不同的分数? |
1 =1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7 =(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)+1/7 =1/2+1/6+1/12+1/20+1/30+1/42+1/7 1 =1/2+1/3+1/6 =1/2+1/3+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10 =1/2+1/3+(1/6-1/7)+(1/7-1/8)+(1/8-1/9)+(1/9-1/10)+1/10 =1/2+1/3+1/42+1/56+1/72+1/90+1/10 1 =1-1/2+1/2-1/3+1/3-1/4+1/4 =1/2+1/6+1/12+1/4 =1/2+1/4+1/6+1/12-1/13+1/13-1/14+1/14-1/15+1/15 =1/2+1/4+1/6+1/156+1/182+1/210+1/15 ....... |
Thanks. Do you think "product of two consecutive numbers" 个分子为1而分母不同的分数 has no solution? "用这个办法是否可以写出和为1的任何多(not a product of two consecutive numbers)个分子为1而分母不同的分数" - I believe so. |
I mean use that exact method we can at least conclude:"用这个办法是否可以写出和为1的任何多(not a product of two consecutive numbers)个分子为1而分母不同的分数" . But I don't think "product of two consecutive numbers" 个分子为1而分母不同的分数 has no solution. For example: 6=2*3 1 =1-1/2+1/2-1/3+1/3 =1/2+1/6+1/3 =1/2+1/3+1/6-1/7+1/7-1/8+1/8-1/9+1/9 =1/2+1/3+1/42+1/56+1/72+1/9 Can we prove there is always solutions for any n>2? |
1 = 1/2 + 1/3 + 1/6 1 = 1/2 + 1/3 + 1/6 - 1/7 + 1/7 = 1/2 + 1/3 + 1/7 + 1/42 1 = 1/2 + 1/3 + 1/7 + 1/42 - 1/43 + 1/43 = 1/2 + 1/3 + 1/7 + 1/43 + 1/(42*43) ...... |
Not long time ago. The difference is, that time the problem wanted 10 fractions, now 7. |
That board has blocked 野 菜 花, which is very unfair. Thank you for the posts. |
If she wants to come back, I would tell the site admin to solve the problem. However, she is happy to be here only. You can find my previous posts here in which we discussed this issue. |
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