The solution is two use a nested for loop:
int startingNum = 1;
int maxNum = 10;
for(int currentNum = startingNum; currentNum <= maxNum; currentNum++)
{
for(int j = 0; j < currentNum; j++)
{
printf("%d", currentNum);
}
printf("\n");
}
#include
#define N 9
//Replace 9 with however high you want it to count
int main()
{
int i, j;
for(i=1;i<=N;i++)
{
for(j=0;j{
printf("%d",i);
}
printf("\n");
}
return 0;
}
int x = 0;
while (x++<5) {
for (int i=0; i<x; ++i) printf ("%d", x);
printf("\n");
} // end while
Output:
1
22
333
4444
55555
#include<stdio.h>
#include<conio.h>
void main()
{
int n,i,j;
clrscr();
printf("enter the last limit:");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
{
printf("%d",i);
}
}
getch();
}
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; }
You can do this with a nested loop: public static void main(String[] args){ for(int i=5, i>0, i--){ for(int j=i, j>0; j--) System.out.print(i); System.out.print(' '); } } The outer loop decrements the integer that you are printing and the inner loop will print the integer the number of times equal to its value.
using System;namespace RightAngleTraingle{class Program{static void Main(string[] args){for (int x = 1; x
55555 55555
4222 + 55555 = 59777
55555=5x41x271 So it is not a perfect square. Square root of 55555 is approximately 235.70108188126758 Dr. Chuck
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"; } }
987598766
55,555
55555
It doesn't.
4938172840
No, it is divisible by 5.
55555
72.27