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 2007-10-08, 03:45 #1 devarajkandadai     May 2004 31610 Posts A puzzle dedicated to Mally Mally was primarily a "puzzles" personality- he used to love solving puzzzles.The following small puzzle is dedicated to him: The function 3^n-2 generates numbers such that their digits add up to 7. a)Is there a simple algebraic explanation?b)Are there other similar functions? Devaraj
 2007-10-08, 04:27 #2 Wacky     Jun 2003 The Texas Hill Country 32·112 Posts Unfortunately, you left out the fact that it fails for n < 2. For larger n, it is very easy to prove.
2007-10-08, 04:50   #3
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by Wacky Unfortunately, you left out the fact that it fails for n < 2. For larger n, it is very easy to prove.
and you both left out that it fails for n = 3 and n > 4

although if we consider D(x) to be the sum of digits of x and D_k(x) to be the k-th iteration of D(x), then lim(k->oo) D_k(3^n-2) = 7 is true for n > 1

2007-10-08, 08:02   #4
Wacky

Jun 2003
The Texas Hill Country

32·112 Posts

Quote:
 Originally Posted by Orgasmic Troll it fails for n = 3
3^3-2 = 27-2 = 25; 2+5=7

Why do you claim that it fails?

 2007-10-08, 08:11 #5 Kees     Dec 2005 22×72 Posts 3^4-2=79...sums to 16, but I suppose you need to keep summing
2007-10-08, 11:22   #6
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by Wacky 3^3-2 = 27-2 = 25; 2+5=7 Why do you claim that it fails?
I meant to say n = 4 and n > 5, I was posting far too late last night

Last fiddled with by Orgasmic Troll on 2007-10-08 at 11:23

 2007-10-08, 11:37 #7 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts Irony It would be ironic if a problem dedicated to Mally were badly formulated
2007-10-08, 11:39   #8
Orgasmic Troll
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Jul 2003

641 Posts

Quote:
 Originally Posted by devarajkandadai Mally was primarily a "puzzles" personality- he used to love solving puzzzles.The following small puzzle is dedicated to him: The function 3^n-2 generates numbers such that their digits add up to 7. a)Is there a simple algebraic explanation?b)Are there other similar functions? Devaraj

for n>1, 3^n is divisible by 9. In a base-r number system, numbers divisible by (r-1) will have an eventual digit sum of (r-1), so in base 10, numbers divisible by 9 have an eventual digit sum of 9. Subtract 2, and you get 7

Similar functions will probably use the same trick. 3^n-5, for example, will always have an eventual digit sum of 4. 18^n-7 will have an eventual digit sum of 2.

multiples of other digits cycle, for example, for n = 1, 2, ... the eventual digit sums of 7n are 7, 5, 3, 1, 8, 6, 4, 2, 9, 7, 5, 3, 1, 8, 6, 4, 2, 9, ...

note, this is a cycle of length 9, so if we wanted to pick out one digit and make it consistent, then we replace n with 9n + k, but then we get 63n + 7k, and the divisibility by 9 trick comes in.

so, basically, here's how you create one of these bad boys. Find some function f(x) where 9|f(x) and then create g(x) = f(x) + k. I would be curious to see an example that doesn't fit this form, or a proof that all of them have to

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