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();
}
}
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"; } }
#include <stdio.h> int main (void) { puts ("1 22 333 4444 55555"); return 0; }
using System;namespace RightAngleTraingle{class Program{static void Main(string[] args){for (int x = 1; x
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<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(); }
55555 55555
4222 + 55555 = 59777
55555=5x41x271 So it is not a perfect square. Square root of 55555 is approximately 235.70108188126758 Dr. Chuck
Closer than it looks
987598766
55,555
55555
It doesn't.
4938172840
No, it is divisible by 5.
55555
72.27