Hey, so I basically have to find all of the Armstrong number from 1 to 1000.

Here's my code so far.


It works, well kind of, I get 1, 153, 370, 371 and 407.
The thing is that I don't get why im not getting the numbers 2-3-4-5-6-7-8-9.
It's probably just a small fix but I don't know.

I don't understand your issue. The Armstrong numbers in the range 1-1000 are 1, 153, 370, 371 and 407. Which is what you are getting. What exactly is the problem?

The Armstrong numbers in the range of 1-1000 are 1-2-3-4-5-6-7-8-9-153-370-371-407.

You can check on the wiki.

Prove it.

I'll get you started

2*2*2 != 2
3*3*3 != 3
4*4*4 != 4


Continue. Show me how 2,3,4,5,6,7,8,9 are Armstrong numbers.

/edit: nevermind, didn't see minko's post

This post was edited by carteblanche on Oct 29 2014 09:49pm

If you are looking for a general Armstrong number that is a number whose digits are raised to the power of the number of digits, then 2,3,4,5,6,7,8,9 would be included, but if you are looking for Armstrong numbers of 3 digits, as you are representing by your code, then they would not be included.

Jesus, I ask for help and I get an arrogant dude answering me with something that doesn't help at all. If im wrong, just say it politely.



Well , I guess you understood.
Alright , so could I just do a IF the number has 2 digits, or 1 in that case and do like : sum + = t*t for 2 number digits and just sum+=t and that would work? Imma try it out.

This post was edited by Jibbosh on Oct 29 2014 09:58pm

I am not being arrogant, and I am not being patronizing. I am using your code to point out your flaws. If I was arrogant, I wouldn't bother helping you at all. The point of my questioning was to get you to think about what your code is doing. Your code is assuming each number is 3 digits, which if you treated 0 as 000, 1 as 001, 2 as 002, etc then your code would be correct. But that is clearly not what you want. My other posts were intended to get you to think about the choices you made when summing up your digits and stumble upon the fact that you are treating each number as a 3 digit, when that is not the case.

Now, back to the question at hand, no. Don't do that. You dont want to have 3 if statements to determine how many digits the number has. Because if you have the data to calculate how many digits the number has then the if statements are not needed. Instead, calculate how many digits your number has, do your logic in your while loop to parse each individual digit, and once you have each digit raise it to the power of the number of digits you calculated and add it to your sum.


Edit: Because you felt insulted by me, I will throw you a bone. Here is a trick to find the number of digits in an integer using Strings



This post was edited by Minkomonster on Oct 29 2014 10:08pm

Didn't feel insulted, it was just a bit rude imo, you probably didn't mean it but that's how I felt it, thanks anyways, I understand now.

Nvm, just saw the int string value conversion in the second line, im tired, lol.

This post was edited by Jibbosh on Oct 29 2014 10:21pm

No, its an integer. The String I was refering to is the the return of String.valueOf(num), which would return "100". Calling the length method on this string returns an integer representing the length of the string.

To be more clear