REMPATTERNS (not necessary)
CLS
FOR I = 1 TO 5 STEP +1
FOR J = 1 TO I STEP+1
PRINT J
NEXT J
NEXT I
END
Hope you find this helpful :)
int main() { int i,j,sum,k; for(i=1;i<=5;i++) { k=1; sum =0; for(j=1;j<=i;j++) { sum = sum+(i*k); k=k*10; } cout<< sum; cout<< "\n"; } }
Implement this method: public static void makeTriangle(int limit) { int count = 0; for(int i = 1; i <= limit; i++) { count = i; while(count > 0) { System.out.print(i); count--; } System.out.println(); } }
#include <stdio.h> int main (void) { puts ("1 22 333 4444 55555"); return 0; }
#include<math.h> main() { int s=1,n,x,i; clrscr(); printf("enter value of n"); scanf("%d",&n); printf("enter value of x"); scanf("%d",&x); for(i=1;i<=n;i++) { s=pow(x,i); } printf("sum of series=%d",s); getch(); }
using System;namespace RightAngleTraingle{class Program{static void Main(string[] args){for (int x = 1; x
Each digit is repeated the number of times equal to its value. 1 22 333 4444 55555 666666 7777777 88888888 999999999
This is one of possible solutions:
int main() { int i,j,sum,k; for(i=1;i<=5;i++) { k=1; sum =0; for(j=1;j<=i;j++) { sum = sum+(i*k); k=k*10; } cout<< sum; cout<< "\n"; } }
To achieve the output "22 4444 666666" in VB (Visual Basic), you can use the Console.WriteLine method. You would write the following code: Console.WriteLine("22 4444 666666") This code directly prints the desired string to the console. If you need to format or generate the numbers dynamically, you can concatenate them accordingly before printing.
The Roman numeral for 4444 is MMMMCDXLIV. The square root of 4444 is 66.663.
Implement this method: public static void makeTriangle(int limit) { int count = 0; for(int i = 1; i <= limit; i++) { count = i; while(count > 0) { System.out.print(i); count--; } System.out.println(); } }
1000000 + 4444 = 1,004,444
500000 + 4444 = 504444
3333 times 4444 = 14811852
4444 - 40% = 2,666.4
13.3453
55 + 4444 - 3 = 4496