Armstrong number in java

I am still somewhat of a beginner to Java, but I need help with my code. I wanted to write an Armstrong Number checker. An Armstrong number is one whose sum of digits raised to the power three equals the number itself. 371, for example, is an Armstrong number because 3^3 + 7^3 + 1^3 = 371. If I understand this concept correctly, then my code should work fine, but I don't know where I made mistakes. I would appreciate if you could help correct my mistakes, but still kind of stick with my solution to the problem, unless my try is completely wrong or most of it needs to change. Here is the code: public class ArmstrongChecker For my tries I used '153' as an input. Thank you for your help! You aren't summing the digits, but the numeric values of the characters representing them. You can convert such a character to its numeric value by subtracting the character '0': int result = 0; for(int i = 0; i < array.length; i++) There is a small logical mistake in your code, You're not converting the character to an integer instead you're doing something like Math.pow('1', 3) -> Math.pow(49, 3) // what you're doing Math.pow(1, 3) // what should be done You should first convert the character to the string using any method below result = (int) Math.pow(array[0],indices); for(int i = 1;i= 0 you can do it like this. Print all the Armstrong numbers less than 10_000. for (int i = 0; i = 0"); } int temp = v; int power = (int)Math.log10(temp)+1; int sum = 0; whi...

I am new to java programming ,please tell me what is wrong with this implementation of the Armstrong's Number Algorithm. It's printing "1" infinite times. An Armstrong number of three digits is an integer such that the sum of the cubes of its digits is equal to the number itself. For example, 371 is an Armstrong number since 3**3 + 7**3 + 1**3 = 371. Reference: Code is here: import java.util.Scanner; public class Program while(num>0)

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Following is my code for getting a three digit public class test I am trying to get any digit Armstrong based on users input, but I am ending up writing too many loops and excess code. The following code check if the scanned number (on console) is a Armstrong number. I've tested it, it works fine. TEST IN CONSOLE import java.util.Scanner; class ArmstrongNumber just change max value of arm_num you will get Armstrong number between 0 to arm_num for(int arm_num = 0 ; arm_num < 100000 ; arm_num++) Please try using below method public static Boolean isArmstrongNumber(int number) Using Java8, you could create a IntStream to generate the Armstrong Numbers. The first step could be to create a function to validate if a number is a Armstrong number: public class ArmstrongNumbersGenerator Example usage: System.out.println(Arrays.toString(ArmstrongNumbersGenerator.inRange(0, 10_000))); Output: [0, 1, 153, 370, 371, 407] System.out.println(Arrays.toString(ArmstrongNumbersGenerator.get(200, 2))); Output: [370, 371] I had written a program for getting all Armstrong Numbers between 1 and 10_000_000 which I had gotten reviewed on Code Review. Following is a minimalist implementation for finding the Armstrong Number using IntStream.range(1, 10_000_000) .filter((n) -> 6161598.59 ns; ?=29084.33 ns @ 3 trials benchmark ms linear runtime Parallel 1.93 ========= Normal 6.16 ============================== Try This code, U can even ask user to input the number to test for Armstrong. class Te...

What is the Armstrong Number? A given number x is called an Armstrong number of order n if its equal to sum of its digits raised to power n, i.e: abc= pow(a,n) + pow(b,n) + pow(c,n) A positive integer of n digits is an Armstrong number (of order 3) if it is equal to the sum of cubes of its digits. For example: Input: 153 Output: Yes 1*1*1 + 5*5*5 + 3*3*3 = 153 Thus, 153 is an Armstrong number Input: 121 Output: No 1*1*1 + 2*2*2 + 1*1*1 = 10 Thus, 120 is not a Armstrong number. Prerequisites and Algorithm In this article, we will learn to check whether an input number is an Armstrong number or not. For this, we will be using a while loop and for • Java if-else statement • Java for Loop • Java while and do-while loop Algorithm: • Take an integer variable x • Value is assigned to the variable x • The digits of the value are split • The cube value of each digit is found • Add the values of all the cubes • The output is saved to sum variable • If sum=X, print Armstrong number If sum != X, print not Armstrong number Program to Check Armstrong Number in Java 1. Armstrong number in Java of order 3 Code: public class Armstrong Output: Explanation: In this program, each number between the given interval low and high, are checked. Post each check, the sum and number of digits are restored to 0. Conclusion In this article, we covered an Armstrong number in java in detail. We saw how to check if a number is an Armstrong number of order n and order 3. We also saw how you could get all t...

A positive integer is called an Armstrong number of order n if abcd... = a n + b n + c n + d n + ... In case of an Armstrong number of 3 digits, the sum of cubes of each digit is equal to the number itself. For example: 153 = 1*1*1 + 5*5*5 + 3*3*3 // 153 is an Armstrong number. This program is built on the concept of Example: Armstrong Numbers Between Two Integers class Main Output 1634 8208 9474 54748 92727 93084 In the above program, each number between the given interval high and low is checked. After each check, the number of digits and the sum result is restored to 0.

In this article, We shall learn what a Java Armstrong Number is. The method for determining if a number is Armstrong or not will be shown, along with its code. We recommend the following topics of • The basic syntax of Java. • Loops in Java. • Decision statement in Java. What is an Armstrong Number in Java? In Java, a number is said to be an Armstrong number if the sum of its own digits raised to a power of digits results in the number itself. For instance, the Armstrong numbers 0, 1, 153, 370, 371, 407, 1634, 8208, and 9474 are both three- and four-digit Armstrong numbers. They are numerous. Consider the number 407, which may be stated as follows because it has three digits: (407) —> (4 * 4 * 4) + (0 * 0 * 0) + (7 * 7 * 7) 64 + 0 + 343 => 407 As 407 is a sum of the cube of its digits, therefore it is an Armstrong number. Implementation of Armstrong Number Program in Java First, we have to declare the class AmstrongNum.Then declare the variables. Create an object of the scanner class as sc and read the number from the user into the variable num. Assign originalNum=num. Find the sum of cubes of digits of num by using a while loop with condition num != 0, • Take each digit by finding the remainder of num divided by 10, digit=num • Find the cube of the digit using the pow() method of Java, add it with cube_sum • To find the next digit we’ll assign the num variable = num/10. After exiting from the loop, check if cube_sum is the same as originalNum, if yes then display the numb...