for (int i = 2; i < 10; i ++) printf("%d\n", i); You did say even and odd numbers between 1 and 10. That's allnumbers between 1 and 10.
Start print "the sum of all even numbers is infinite" end
Program: include "emu8086.inc" ORG 100h MOV AX, 0004H //Move 1st 16-bit number to AX. MOV BX, FFFEH //Move 2nd 16-bit number to BX. IMUL BX //Multiply BX with AX and the result will be in DX:AX. CALL PRINT_NUM //Print the result. RET //Return. DEFINE_PRINT_NUM //Declare function. END
#include<stdio.h> #include<conio.h> void main() { clrscr(); for (int i=1;i<=100;i++) { if(i%2 !=0 && i%3 !=0) { printf("\n Numbers that are not divisible by 2 or 3:%d \n ",i); } printf("\n"); } getch(); }
Since you didn't specify, I will give you the answer for a TI 83 Plus Silver Addition :Input A :Input B :ClrHome <--found in the prgm menu with input :Output(1,1,A+B) <--Output also found in same menu as ClrHome :Output(4,1,A*B)
For N = 2 to 30 STEP 2 Sum = Sum + N Next N Print "The sum is "; Sum; ". Have a nice day. Come back and see us." END
You can use int i; for (i = 10; i <= 50; i += 2) {//print i} as a program to print even numbers between 10 and 50.
Q.1 Write a program to print first ten odd natural numbers. Q.2 Write a program to input a number. Print their table. Q.3 Write a function to print a factorial value.
#include
Write a c program to print the 100 to 1 nos
you do this 10 print "0112358132134" use the whole of the thing
write a program to print A to Z on screen in c?
Try the triangle program on a search engine. Replace numbers with stars and that should do the trick
echo 'print a pattern'
Cls input "enter two no.s ",a,b sum=a+b print "sum = ";sum end
# include<stdio.h> void main() { int i = 1; while(i<11) { if(i%2==1) { printf("\n%d",i); } i++; } }
#include<stdio.h> #include<conio.h> void main() { int i,j; for(i=51;i<100;i=i+2) { printf("%d", i); } getch(); }
#include <iostream> int main() { for(int i=51; i <= 100; i+=2) { cout << i << endl; } return 0; }