A puzzle dedicated to Mally

[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

[Wacky]

Unfortunately, you left out the fact that it fails for n < 2.

For larger n, it is very easy to prove.

[Orgasmic Troll]

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

[Wacky]

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?

[Kees]

3^4-2=79...sums to 16, but I suppose you need to keep summing

[Orgasmic Troll]

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

[davieddy]

It would be ironic if a problem dedicated to Mally were
badly formulated

[Orgasmic Troll]

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