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
To print the given sequence in Java, you can use a loop to iterate through each number and print it the corresponding number of times. Here's an example code snippet to achieve this: public class NumberSequence { public static void main(String[] args) { int[] numbers = {5, 4, 3, 2, 1}; for (int num : numbers) { for (int i = num; i >= 1; i--) { System.out.print(num); } System.out.print(" "); } } } This code will output: 55555 4444 333 22 1
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"); }
55555 55555
4222 + 55555 = 59777
55555=5x41x271 So it is not a perfect square. Square root of 55555 is approximately 235.70108188126758 Dr. Chuck
987598766
55,555
It doesn't.
55555
4938172840
No, it is divisible by 5.
72.27
88,944,443
55555