Hello can u tell me what's wrong?

Question about the task

Sum of the digits of a three-digit number

Under discussion

public class Solution { public static void main(String[] args) { System.out.println(sumDigitsInNumber(546)); } public static int sumDigitsInNumber(int number) { int number1 = (5+4+6); return number1; } }

Write the code for sumDigitsInNumber(int number). The method takes a three-digit whole number. You need to calculate the sum of the digits of this number, and then return the result.

Consider this example:
The sumDigitsInNumber method is called with the argument 546.

Example output:
15

Requirements:

The program must not read data from the keyboard.
The sumDigitsInNumber(int) method must be public and static.
The sumDigitsInNumber method must return an int.
The sumDigitsInNumber method should not display anything.
The sumDigitsInNumber method must return the sum of all digits in the parameter number.

package com.codegym.task.task01.task0132; /* Sum of the digits of a three-digit number */ public class Solution { public static void main(String[] args) { System.out.println(sumDigitsInNumber(546)); } public static int sumDigitsInNumber(int number) { int number1 = 5; int number2 = 4; int number3 = 6; return number1+number2+number3; } }

Comments (5)

Popular
New
Old

You must be signed in to leave a comment

null

Level 13 , Jin'e, 中国

19 July 2019, 04:19

You haven't use the parameter, you can get each bit of the value by "div" or "mod".And then you plus this num.Hope this can help you.

William Hoang

Level 10 , Hanoi, Viet Nam

17 July 2019, 18:13

on the method sumDigitisNumber. It should use "parameter" i see your method do not use "parameter" your parameter on this case is 546. That mean this method should separate 546 to three numbers are 5,4,6 then plus its to 15. Do not create 3 numbers 5,4,6 and plus its to 15

Bazuka

Level 19 , Krakow, Poland

19 July 2019, 08:25

William Hoang

Level 10 , Hanoi, Viet Nam

19 July 2019, 14:59

int = number1 = (5+4+6)

your code didn't use parameter :D let's try this.

public static int sumDigitsInNumber(int number) {
int number1 = number%10;  // get number 6 from parameter - number
int number2 = (number%100 - number1) / 10; // get number 4 from parameter
int number3 = (number - number2*10 - number1 ) / 100; // get number 5 from patemeter
return number1+number2+number3;}

parameter on this case is 546. When you type

System.out.println(sumDigitsInNumber(546));

It should show 15. But when you type

System.out.println(sumDigitsInNumber(123));

it will show 6. On your code, you didn't use parameter. So when you type

System.out.println(sumDigitsInNumber(546));

it shows 15, it maybe correct. But when you type

System.out.println(sumDigitsInNumber(123));

your result still shows 15 and incorrect :D it should be 6.

William Hoang

Level 10 , Hanoi, Viet Nam

19 July 2019, 15:06

int b = sumDigitsInNumber(parameter)

call method and put method to work with parameter

public static int sumDigitsInNumber(int parameter)
a = parameter*2;
return a

double value of parameter and assgin b