Use loop structures below to print the even numbers 2 to 100
Solution
#include<iostream.h>
#include<cono.h>
void main()
{
int i;
for(i=1;i<=100;i++)
{
if(i%2==0)
cout<<i<<endl;
}
}
A.for(i=1;i<=100;i=i+2)
{
if(i%2==0)
cout<<i<<endl;
}
B.
for(i=1;i<=50;i++)
{
if(i%2==0)
cout<<i<<endl;
}
C.
for(i=100;i>=0;i--)
{
if(i%2==0)
cout<<i<<endl;
}
d.
main()
{
int t ;
for ( ; ; ) {
scanf(\"%d\" , &t) ;
if ( t==10 ) break ;
}
E.
main()
{
int t ;
while ( ; ; ) {
scanf(\"%d\" , &t) ;
if ( t==10 ) break ;
}
F:
main()
{
int t ;
do {
scanf(\"%d\" , &t) ;
if ( t==10 ) break ;
}while(;;)
2.void main()
{
int due=0,rec=0;
cout<<\"enter the amount:\";
cin>>rec;
cout<<\"Enter the Due amount:\";
cin>>due;
cout<<\"Amout:<<\"$\"<<rec
}
3.
 #include<stdio.h>
 #include<conio.h>
 void main()
 {
 int n;
 clrscr();
 printf(\"Month No:-> \");
 scanf(\"%d\",&n);
 switch(n)
 {
    case 1:
    case 3:
    case 5:
    case 7:
    case 8:
    case 10:
    case 12:
    printf(\"Month have 31 days \ \");
    break;
    case 2:
    printf(\"2nd month is a february and have 28 days \ \");
    printf(\"in leap year The Feb Have 29 days\ \");
    break;
    case 4:
    case 6:
    case 9:
    case 11:
    printf(\"Month have 30 days \ \");
    break;
 default:
    printf(\"invalid Month number\ Please try again ....\ \");
    break;
 }
 getch();
 }



