一至七十里,与七十互质的正整数有多少
答案:3 悬赏:30
解决时间 2021-03-12 17:06
- 提问者网友:一人心
- 2021-03-12 02:23
一至七十里,与七十互质的正整数有多少
最佳答案
- 二级知识专家网友:24K纯糖
- 2021-03-12 03:04
1至70中,奇数共70÷2=35个
除去1之外的奇数整数共35-1=34个
奇数中,5的倍数共35÷5=7个
7的倍数共35÷7=5个
即是7的倍数又是5的倍数共35÷35=1个
∴与70互质的正整数共 34 - (7+5-1) = 34 - 11 = 23个
除去1之外的奇数整数共35-1=34个
奇数中,5的倍数共35÷5=7个
7的倍数共35÷7=5个
即是7的倍数又是5的倍数共35÷35=1个
∴与70互质的正整数共 34 - (7+5-1) = 34 - 11 = 23个
全部回答
- 1楼网友:瘾与深巷
- 2021-03-12 03:37
1,3,9,11,13,17,19,23,27,29,31,33,37,39,41,43,47,51,53,57,59,61,67,69,共24个
- 2楼网友:我的任性你不懂
- 2021-03-12 03:31
private sub command1_click() n = cint(inputbox("请输入n:")) t = n for i = 2 to n / 2 p = t do while t mod i = 0 t = t \ i loop if p <> t then s = s & i & " " next i msgbox n & "的所有质因子是:" & s end sub
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算法二:
private function sushu(a as integer) as boolean for i = 2 to cint(sqr(a)) if a mod i = 0 then sushu = false exit function end if next i sushu = true end function
private sub command1_click() dim s as string, n as integer, i as integer n = cint(inputbox("请输入n:")) s = "" for i = 2 to cint(sqr(n)) if n mod i = 0 and sushu(i) = true then s = s & i & " " end if next i msgbox n & "的所有质因子是:" & s end sub
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