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#1 2005-08-04 22:50:35

wcy
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Digit-perfect Numbers!!

I found some numbers that are really interesting!

n:number of digits

n=3,
1^3+5^3+3^3=153
3^3+7^3+0^3=370
3^3+7^3+1^3=371
4^3+0^3+7^3=407

n=4,
8^4+2^4+0^4+8^4=8208
9^4+4^4+7^4+4^4=9474
1^4+6^4+3^4+4^4=1634

n=5,
92727
54748
93084

n=6,
548834

n=7,
9926315

Last edited by wcy (2012-01-14 17:28:33)

 

#2 2005-08-04 22:56:46

wcy
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Re: Digit-perfect Numbers!!

-

Last edited by wcy (2012-01-14 17:29:06)

 

#3 2005-08-05 00:30:42

mathsyperson
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Re: Digit-perfect Numbers!!

I don't know if these two count because the power isn't the number of digits, but 4150= 4^5+1^5+5^5+0^5 and 4151= 4^5+1^5+5^5+1^5

There are a few more where n=7: 1,741,725; 4,210,818; 9,800,817
There's an n=8: 24,678,050
n=9: 146,511,208; 472,335,975; 534,494,836; 912,985,153
n=10: 4,679,307,774

A bit different: 3435= 3^3+4^4+3^3+5^5 and 40585= 4!+0!+5!+8!+5!

And for the grand finale, wait, this is too amazing to just put here. I'll put it in this

Last edited by mathsyperson (2005-08-05 00:46:17)


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It wanted to be normal.
 

#4 2005-08-05 08:58:23

MathsIsFun
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Re: Digit-perfect Numbers!!

mathsy! How did you do it!

I was thinking along the lines:

Start with: 1^3+5^3+3^3=153
Expand decimal: 1^3+5^3+3^3=110^2 + 510^1 + 310^0

Algebraically: a^3+b^3+b^3=a10^2 + b10^1 + c10^0

But that was as far as I got ...


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#5 2005-08-05 14:17:43

ganesh
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Re: Digit-perfect Numbers!!

Superlative work, Mathsy smile
How did you do it?


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#6 2005-08-05 18:01:56

mathsyperson
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Re: Digit-perfect Numbers!!

Sorry to disappoint everyone, but I kind of cheated...


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It wanted to be normal.
 

#7 2005-08-05 18:05:36

justlookingforthemoment
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Re: Digit-perfect Numbers!!

Where did you cheat? I can see the website, but not your answers...

 

#8 2005-08-05 18:06:47

insomnia
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Re: Digit-perfect Numbers!!

CHEATING!!!!

How dare you!!!

Even though we all do it.


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#9 2005-08-05 18:10:13

mathsyperson
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Re: Digit-perfect Numbers!!

Scroll down to just below halfway and they're in a big black box.


Why did the vector cross the road?
It wanted to be normal.
 

#10 2005-08-05 18:11:39

justlookingforthemoment
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Re: Digit-perfect Numbers!!

Oh right. I see. Tut tut, Mathsy. You shouldn't have told us...

 

#11 2005-08-05 18:46:53

insomnia
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Re: Digit-perfect Numbers!!

Bad boy!!


Friends are angels who lift our feet when our own wings have trouble remembering how to fly
 

#12 2005-08-05 20:09:38

MathsIsFun
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Re: Digit-perfect Numbers!!

Ummm ... mathsy was not cheating when he tells us, right?

It comes under the heading "havin' a little fun with us" I think.

Good find, by the way. Interesting website, too.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#13 2005-08-05 21:25:39

wcy
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Re: Digit-perfect Numbers!!

wow this is cool!!

i was stuck at n=8, where there were 100 million possibilities.. i was using Javascript, and my computer sort of died..

now for the ultimate question:

Prove, using induction or otherwise, that there are solutions for all n element of Z+, n>2.

(i don't know the answer myself smile )

Last edited by wcy (2005-08-05 21:26:14)

 

#14 2005-08-05 21:54:13

MathsIsFun
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Re: Digit-perfect Numbers!!

For a computer-based solution you may be better off using a faster language like C, Java or .Net. But before that, there may be some tricks to speed it up - for example, there may be whole ranges that can be excluded from the search. (But I didn't look at your code, so maybe you are already doing that.)

So, are we back to a^3 + b^3 + c^3 = a10^2 + b10^1 + c10^0 ?

This can be rearranged to: a^3 + b^3 + c^3 - a10^2 - b10^1 - c10^0 = 0
Then: a^3 - a10^2 + b^3 - b10^1 + c^3 - c10^0 = 0
Then: a(a^2 - 10^2) + b(b^2 - 10^1) + c(c^2 - 10^0) = 0

Is this helping or making it worse I wonder.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman
 

#15 2005-08-05 21:58:09

ganesh
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Re: Digit-perfect Numbers!!

That is, it is to be proved that
(abcd..........)^n = a^n+b^n+c^n+d^n..................
Here, both a,b,c,d...etc and n go on and on.
It is easy to say there would always exist a n digit number whose sum of digits raised to the power n is equal to the number. To prove that may be quite difficult. As n becomes higher, the combinations available increase, thereby increasing the possibility of such a number existing. smile


Character is who you are when no one is looking.
 

#16 2005-08-05 22:58:29

mathsyperson
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Re: Digit-perfect Numbers!!

I think that as there is no answer for n=2, there are probably other values of n for which there are no answers. For this reason, I think it would be better to prove by counter-example. Of course, we'll need 10^n counter-examples, but it's still easier than proving something that's not true.

If you still want to try, then don't be discouraged by my scepticism. After all, Fermat's Last Theorem doesn't work with n=2 either!


Why did the vector cross the road?
It wanted to be normal.
 

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