编写函数判断一个数能否同时被3和5整除

#includevoidmain(){intn;printf("input:\n");scanf("%d",&n);if(n%3==0&&n%5==0)printf("yes\n");elseprin

被3整除的数,各个数字和为3倍数.被5整除的数,末位数字为0或5.==>同时被3和5整除的数,各个数字和为3倍数且末位数字为0或5.或者更简单些,能被15整除.算法：1、判断是否为3倍数————把n拆

if(number%3==0&&number%5==0){System.out.println("number"+能被三和五整除);}else{System.out.println("number"+

#includeintmain(intargc,char*argv[]){intnum;printf("输入一整数:");scanf("%d",&num);if(num%3==0&&num%5==0&

PrivateFunctionjo(ByValnAsInteger)n=InputBox("请输入一个数字")IfnMod20ThenMsgBoxn&"是奇数"ElseMsgBoxn&"是偶数"End

#include"stdio.h"main(){intx;printf("请输入一个数字:");scanf("%d",&x);if(x%3==0)printf("该数字能被3整除.");if(x%5=

#includemain(){intn;printf("Pleaseenteranumber:");scanf("%d",&n);if(n%2==0)printf("Thenumber%dis偶数\n

intwanshu(intx){inti,sum;for(i=1,sum=0;i

DimnAsIntegerPrivateSubForm_Click()Ifn/2=0ThenPrint"n是偶数"ElsePrint"n是奇数"EndIfEndSubPrivateSubForm_Lo

'判断一个数m是否是完数,是则返回True,否则返回FalseFunctionTestWanShu(ByValmasLong)AsBooleanDimSumAsLongDimiAsLongIfm

if(x%3==0)y=3;if(x%5==0)y=5;if(x%7==0)y=7;if(x%7==0&&x%5==0&&x%3==0)y=15;if(x%7==0&&x%5==0)y=12;if(x

3和5的最小公倍数为15所以只要一个数能够被15整除,就能同时被3和5整除

以下是关键代码intnum;//这个是要你判断的数if(num%3==0&&num%5==0)//判断能否被整除printf("能被3和5整除");

我用C语言编一下啊.其他语言类似!#includevoidsort(intn){if(n%2==0)printf("thenumberisdouble\n");elseprintf("thenumbe

voidmain(){intn;printf("请输入一个整数:");scanf("%d",&n);if(n%3==0&&n%5==0&&n%7==0)printf("\n此数能被3,5,7整除\n"

oolfun(inti)//判断i{if(i%3==0)returntrue;returnfalse;}

DimnumDimstrstr=""Fornum=1000To2000str=str+Cal(num)NextMsgBoxstrFunctionCal(num)IfnumMod17=0AndnumMo

#includevoidmain(){intn;printf("pleaseinputthedata:\n");scanf("%d",&n);if(n%3==0&&n%5==0)printf("the

intsum(intn){intx;If(x%3==0&&x%5==0)sum(n)==1;elsesum(n)==0;}这样的代码里边sum(n)成了递归了,当然死循环了#include"stdio

clearinput"请输入一个数："toxifx%3=0andx%5=0?"能被3和5整除"else?"不能被3和5整除"endif