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Old 2007-10-08, 03:45   #1
devarajkandadai
 
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Default 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
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Old 2007-10-08, 04:27   #2
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Unfortunately, you left out the fact that it fails for n < 2.

For larger n, it is very easy to prove.
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Old 2007-10-08, 04:50   #3
Orgasmic Troll
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Quote:
Originally Posted by Wacky View Post
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
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Old 2007-10-08, 08:02   #4
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Quote:
Originally Posted by Orgasmic Troll View Post
it fails for n = 3
3^3-2 = 27-2 = 25; 2+5=7

Why do you claim that it fails?
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Old 2007-10-08, 08:11   #5
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3^4-2=79...sums to 16, but I suppose you need to keep summing
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Old 2007-10-08, 11:22   #6
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Quote:
Originally Posted by Wacky View Post
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
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Old 2007-10-08, 11:37   #7
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Default Irony

It would be ironic if a problem dedicated to Mally were
badly formulated
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Old 2007-10-08, 11:39   #8
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Quote:
Originally Posted by devarajkandadai View Post
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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