用0,1,2,3....9这10个数字组成6个质数,每个数字至多用一次,每个质数都不大于

2024-12-28 12:05:12
推荐回答(3个)
回答1:

一共有两组答案:
第一组是:2,5,61,409,83,7
第二组是:2,5,67,401,89,3

思考过程如下:
由于素数里面除2外都是奇数,而0~9只能挤出5个奇数,
故,这六个素数中必有素数2;
其他每个奇数都要组成一个素数;
有结论:每个奇数在素数中只能用一次——记为结论1
由于0不能放在数的最高位,
故0必用在103 107 109 307 401 409中(这些数来自楼上);
再由结论1可知:
0必用在401或409中;(可以看出4一定和0出现在一起)(记为结论2)
再考虑5,由于大于5的任何以5结尾的数必是5的倍数,
故5只能单独成为一个素数:
由偶数只剩6,8两个,奇数剩3,7及9或1中的一个~
由于所有素数都小于500,故不可能出现68X,或86X之类的数
即6和8不能成双出现在素数中,而以6开头的素数有61 67;
以8开头的素数有83 89;
若6和1搭配成61则由结论2知素数中必有409,剩下8,3,7只能组合出7和83,得到一个答案!
若6和7搭配成67,则剩下8,3及9或1中的一个,8可以和3或9搭配:
——若8和3搭配,则9或1必有一数单独成素数,但9或1都不是素数,矛盾!
——若8和9搭配,1只能存在于401中,剩下3单独成素数,又一个答案!

故这题有两组答案
第一组是:2,5,61,409,83,7
第二组是:2,5,67,401,89,3

回答2:

2 3 5 7 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 103 107 109 127 137 139 149 157 163 167 173 179 193 197 239 241 251 257 263 269 271 281 283 293 307 317 347 349 359 367 379 389 397 401 409 419 421 431 439 457 461 463 467 479 487 491

500以内的本身数字不重复的质数是这些
然后在用搜索的方法吧
好像不是数学题,倒像是编程题 呵呵

回答3:

2,3,5,7,41,89
2,3,5,7,61,89
2,3,5,7,89,401
2,3,5,7,89,461
2,3,5,7,61,409
2,3,5,47,61,89
2,3,5,41,67,89
2,3,5,67,89,401
2,5,7,43,61,89
2,5,7,61,83,409

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