判断i能否被3和5整除的python表达式为

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procedure TForm1.Edit1KeyPress(Sender: TObject; var Key: Char);

begin

if not (key in [*0*..*9*,*.*,#8]) then

key:=#0;

end;

procedure TForm1.Button1Click(Sender: TObject);

var

n,m:real;

i:string;

begin

i:=Edit1.Text;

n:=strtofloat(i)/3;

m:=strtofloat(i)/5;

if (pos(*.*,Floattostr(n))=0) and (pos(*.*,Floattostr(m))=0) then

showmessage(*该数可以同时被3,5整除*);

if (pos(*.*,Floattostr(n))=0) and (pos(*.*,Floattostr(m))0) then

showmessage(*该数只能被3整除*);

if (pos(*.*,Floattostr(n))0) and (pos(*.*,Floattostr(m))=0) then

showmessage(*该数只能被5整除*);

if (pos(*.*,Floattostr(n))0) and (pos(*.*,Floattostr(m))0) then

showmessage(*该数都不能被3,5整除*);

end;

procedure TForm1.Edit1Change(Sender: TObject);

begin

if strtoint(edit1.Text)>100 then

begin

showmessage(*程计100*);

edit1.Text:=*100*;

end

end;

还有个DVI函数也可以实现 DVI返回result.quot 是商 ， result.rem是余数 可以判断余数为0 去调用