## Problem 023

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phillip1882
Posts: 5
Joined: Sat Jun 21, 2008 10:08 pm

### Problem 023

i believe i solved to problem, but the answer i gave doesn't check. i used basically a three step algorithm to sovle,
first generate all abundant numbers between 1 and 28123, then add every abundant number to every abundant number, crossing off each total from the list of numbers between 1 and 28123, and finally adding all numbers that were not crossed off. i checked by hand the first 10 abundant numbers my program generates, and they are correct, so i'm not sure where i am going wrong, can someone look at my code and maybe give me a hint?

daniel.is.fischer
Posts: 2400
Joined: Sun Sep 02, 2007 11:15 pm
Location: Bremen, Germany

### Re: problem #23

I could take a look, PM the code.
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.

joesmoe10
Posts: 1
Joined: Wed Nov 05, 2008 6:40 pm

### Problem 23 - incorrect info?

Hey I'm working on Problem 23 and I noticed a discrepancy. Problem 23 states
By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
However the Wolfram http://mathworld.wolfram.com/AbundantNumber.html site states:
Every number greater than 20161 can be expressed as a sum of two abundant numbers.
This shouldn't affect an answer but it is curious. Perhaps some new research came out?

stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 11:57 pm
Location: Netherlands

### Re: Problem 23 - incorrect info?

However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is 20161.

ed_r
Posts: 1009
Joined: Sun Jul 29, 2007 10:57 am

### Re: Problem 23 - incorrect info?

I have some sympathy for joesmoe10. I mean, "this upper limit cannot be reduced any further by analysis" ... that's quite a bold claim, isn't it? What about mathematical analysis by cases? - not pretty, but still "analysis".

IMHO, PE#23 should've used the true bound instead of one found in some random guy's proof on the net.
!647 = &8FDF4C

euler
Posts: 3228
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### Re: Problem 23 - incorrect info?

@joesmoe10: It is quite possible that some new research has come out, but I am unaware of any; I'd be happy to be corrected though (see below.)

@ed_r: What do you mean "some random guy's proof on the net"? It's elementary number theory and apart from a couple of papers published: one in the early 1960's and the other in the early 1970's, I am unaware of any reduction in the limit without the aid of a computer. Remember that I put that problem together over six years ago and I think a little latitude is needed when retrospective criticisms are made; access to a whole lot more information is available since then. I am quite happy to alter the limit/wording to reflect the lower limit if that is generally preferred. However, I personally think a clear line lies between the type of mathematical analysis that relies on deduction and the type that relies on exhaustive testing with the aid of a computer.

Perhaps I need to change the word "analysis" to remove/replace ambiguity?

impudens simia et macrologus profundus fabulae

ed_r
Posts: 1009
Joined: Sun Jul 29, 2007 10:57 am

### Re: Problem 23 - incorrect info?

Some random guy's proof = whatever it was that I found by googling your "analysis" bound.

Re latitude - yes, of course. That's why I won't ask you to change it.
!647 = &8FDF4C

Ted
Posts: 19
Joined: Sun Apr 02, 2006 10:46 pm

### Re: Problem 23 - incorrect info?

My vote is to leave it as is. Its one of the many notes to self I've made to explore after I solve all the other problems, as to why analysis can't (or didn't so far) lower the bound.

I am fascinated by what can be brute forced and what can be analyzed, and while working many of Project Euler's puzzles I wonder whether I've uncovered all the "tricks," whhich, more often than not, I haven't.

With Problem 23, at least the literature says, that perhaps no one has!

euler
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### Re: Problem 23 - incorrect info?

For your reference, I've put together a couple of problems on my other website that addresses this proof.
Even Sum Of Two Abundant Numbers (Every even integer n [ge] 48 can be written as a sum of two abundant numbers)
Sum Of Two Abundant Numbers (Every integer greater than 28123 can be written as a sum of two abundant numbers)

Although I am sure that many visitors will find the proof useful, I am a little uncomfortable with the "28123" problem as it is not only somewhat contrived, but it goes beyond the difficulty I like to post on mathschallenge.net; the website is aimed at the range of school students.

impudens simia et macrologus profundus fabulae

DDgeva
Posts: 15
Joined: Mon Dec 29, 2008 12:47 pm

### Re: Problem 023

I also can't seem to get the right answer even though I think my code is correct.
the sum I'm getting is 4248567. am I way off?
one problem I'm getting is that my program states 62 as a number that can't be produced by adding two abundant numbers even though it isn't true. can someone please check my code? it's quite clear (JAVA):

Code: Select all

/* snip */

if it helps, i can upload the list of numbers that my program lists as numbers that can't by expressed by adding two abundants.
Last edited by daniel.is.fischer on Tue Feb 03, 2009 7:24 pm, edited 1 time in total.
Reason: Snip code

quilan
Posts: 182
Joined: Fri Aug 03, 2007 11:08 pm

### Re: Problem 023

Thou shalt not post code. I'd edit your post asap.
ex ~100%'er... until the gf came along.

Tommy137
Posts: 238
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Location: Cologne, Germany
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### Re: Problem 023

@ DDgeva

Check your sumDivisors-function. 62 should be the sum of the two abundant numbers 20 and 42.

hk
Posts: 10712
Joined: Sun Mar 26, 2006 10:34 am
Location: Haren, Netherlands

### Re: Problem 023

Before it was snipped away (thanks Daniel for doing so) I had a quick glance at the code and spotted the problem Tommy presumably spotted too.
If you can't find the problem, you can PM me.

DDgeva
Posts: 15
Joined: Mon Dec 29, 2008 12:47 pm

### Re: Problem 023

quilan wrote:Thou shalt not post code. I'd edit your post asap.
sorry about that.. I forgot this isn't a problem specific thread that only people who solved correctly can see.

I found out what one of my problems was. I was checking if the number % its root = zero, but the root was rounded down so the square root of 20 was considered 4, therefore I was decreasing the sum of 20's divisors by 4. I solved that one, but I still get a wrong answer.

I got the impression that I'm allowed to post wrong solutions, so this is what I get now:
*censured for being right //Haha sounds a bit political

any ideas guys? anyone willing to check my code on PMs?

EDIT: sorry for the trouble, I probably skipped a character or something because now I tried inputting the answer and it's correct.

hk
Posts: 10712
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Location: Haren, Netherlands

### Re: Problem 023

DDgeva wrote:
quilan wrote:Thou shalt not post code. I'd edit your post asap.
sorry about that.. I forgot this isn't a problem specific thread that only people who solved correctly can see.
Wouldn't have been difficult to grasp though, because you did not solve it yet.

Tommy137
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### Re: Problem 023

hk wrote:
DDgeva wrote:
quilan wrote:Thou shalt not post code. I'd edit your post asap.
sorry about that.. I forgot this isn't a problem specific thread that only people who solved correctly can see.
Wouldn't have been difficult to grasp though, because you did not solve it yet.
I'm always astonished by your display of pure logic

hk
Posts: 10712
Joined: Sun Mar 26, 2006 10:34 am
Location: Haren, Netherlands

### Re: Problem 023

Obviously a result of more than 30 years of math teaching.

iamhigh
Posts: 7
Joined: Fri Apr 03, 2009 11:24 am

### Problem 23 Hints

Can somebody verify that the total number of Abundant numbers from 1 to 28123 is 6965?

Thank you!

Tommy137
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### Re: Problem 23 Hints

Yep, that's correct.