20071008, 03:45  #1 
May 2004
316_{10} 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^n2 generates numbers such that their digits add up to 7. a)Is there a simple algebraic explanation?b)Are there other similar functions? Devaraj 
20071008, 04:27  #2 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Unfortunately, you left out the fact that it fails for n < 2.
For larger n, it is very easy to prove. 
20071008, 04:50  #3  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:
although if we consider D(x) to be the sum of digits of x and D_k(x) to be the kth iteration of D(x), then lim(k>oo) D_k(3^n2) = 7 is true for n > 1 

20071008, 08:02  #4 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 

20071008, 08:11  #5 
Dec 2005
2^{2}×7^{2} Posts 
3^42=79...sums to 16, but I suppose you need to keep summing

20071008, 11:22  #6 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
I meant to say n = 4 and n > 5, I was posting far too late last night
Last fiddled with by Orgasmic Troll on 20071008 at 11:23 
20071008, 11:37  #7 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Irony
It would be ironic if a problem dedicated to Mally were
badly formulated 
20071008, 11:39  #8  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:
for n>1, 3^n is divisible by 9. In a baser number system, numbers divisible by (r1) will have an eventual digit sum of (r1), 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^n5, for example, will always have an eventual digit sum of 4. 18^n7 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 9f(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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