Coding Challenge #1

[Blasted heath]

Whee! I now thoughly despise happy numbers, but it's done now, yay!

[ODDin]

I must admit - I didn't code this, since I didn't get the math about the happy numbers.
We aren't even close to this here, in 10th class (we started studying parabulas (sp?) only a month and a half ago, so what do you want?).

And another question - after the challenge is over, will you show the source codes of the winners (if they won't mind, of course)?

[Long Shoota]

What's hard to get about the maths?

You take a number, say 7

You square 7
7^2 = 49

You square each digit of the number
4^2 = 16
9^2 = 81

You add the numbers
16 + 81 = 97

You repeat the proceedure until the number reaches 1 (if it ever does).

[Blasted heath]

I didn't know what happy numbers were either... there's nothing hard about them.

I'll post a link to this script what I wrote here, 'cos there's no way of stealing the code, you filthy theives. (Not that you'd want to anyway...)

http://www.ecowonline.co.uk/misc/happynumbers.php

[NeoThermic]

What's hard to get about the maths?

You take a number, say 7

You square 7
7^2 = 49

You square each digit of the number
4^2 = 16
9^2 = 81

You add the numbers
16 + 81 = 97

You repeat the proceedure until the number reaches 1 (if it ever does).

Er... you're missing something. I think quite a few of you might of. If it never reaches 1, it repeats a number in the series periodically. So you will need to store all the calculations you've done, and check against that as well as against 1. If it repeats without hitting 1, its not a Happy Number.

NeoThermic

[ODDin]

Oh. Then it was just the page I tried that put it in a way too complicated way...
Ok.
But I don't know whether I'll have time to code that in the near future...
Hmm, maybe I'll just pass the 1st challenge then.

[Disco Stuie_uplink]

Forgive me for being dumb (and I really think I am being so) but if you are following the procedure above, of

What's hard to get about the maths?

You take a number, say 7

You square 7
7^2 = 49

You square each digit of the number
4^2 = 16
9^2 = 81

You add the numbers
16 + 81 = 97

You repeat the proceedure until the number reaches 1 (if it ever does).

are you then squaring 9 and squaring 7, then adding 81 to 49; or, are you squaring 97, then squaring the digits of that answer?

And, in any case, I don't quite see how you can end up with 1 at any time - if you are always adding integer squares to each other, you are always going to have to have >2 as an answer (no two integers add to make 1, is what I mean). Explanation?

[Long Shoota]

Square 9 then 7, then add them.

So what about if a number becomes 10? Does that also add to go higher than 2?

[Blasted heath]

Forgive me for being dumb (and I really think I am being so) but if you are following the procedure above, of

What's hard to get about the maths?

You take a number, say 7

You square 7
7^2 = 49

You square each digit of the number
4^2 = 16
9^2 = 81

You add the numbers
16 + 81 = 97

You repeat the proceedure until the number reaches 1 (if it ever does).

are you then squaring 9 and squaring 7, then adding 81 to 49; or, are you squaring 97, then squaring the digits of that answer?

And, in any case, I don't quite see how you can end up with 1 at any time - if you are always adding integer squares to each other, you are always going to have to have >2 as an answer (no two integers add to make 1, is what I mean). Explanation?

Example: 68 is a happy number, its digits squares add up to 100, 1^2 + 0^2 + 0^2 = 1

You are doing the same as you did to the first number to the number it produced.

[Disco Stuie_uplink]

*me blushes*

[Long Shoota]

*me mocks*

[Disco Stuie_uplink]

*me decides that now is not the time to mention that he is doing A-Level Further Maths...*

*me goes and hides in a corner, slowly counting 1 to 10, over and over and over*

[Darksun]

My maths teacher wrote a program to do this, that's why I decided to make this challenge

And Blasted heath, could you email me the source code for your script?

[Blasted heath]

I did yesterday...

[Stewsburntmonkey]

The solution is actually rather trivial in Perl (yay hashes).