 [SOLVED] powers of 2  

Author  Message 

peyman Euclid
Posts : 49 Join date : 20091106 Age : 94 Location : dense in the universe
 Subject: [SOLVED] powers of 2 Thu Nov 19, 2009 7:12 pm  
 I there a power of 2 such that its digits could be rearranged and made into another power of 2? (No zeroes are allowed in the leading digit: for example. 0032 is not allowed.) 

 
Bruno Admin
Posts : 184 Join date : 20090915 Age : 31 Location : the infinite, frictionless plane of uniform density
 Subject: Re: [SOLVED] powers of 2 Thu Nov 19, 2009 10:37 pm  
 No, because obviously the biggest of the two could be at most 8 times the other, and the difference between the two would be divisible by 9 (casting out nines!). But it's easy to check that none of
2^(m+3)2^m,
2^(m+2)2^m,
2^(m+1)2^m
are divisible by 9. 

 
Bruno Admin
Posts : 184 Join date : 20090915 Age : 31 Location : the infinite, frictionless plane of uniform density
 Subject: Re: [SOLVED] powers of 2 Thu Nov 19, 2009 10:39 pm  
 Now that was literally a 60second problem! Mohammad would be proud. 

 
peyman Euclid
Posts : 49 Join date : 20091106 Age : 94 Location : dense in the universe
 Subject: Re: [SOLVED] powers of 2 Thu Nov 19, 2009 11:31 pm  
 A shorter answer would've been "No, obviously!"
Can you provid more details of your "obvious" claims? 

 
Bruno Admin
Posts : 184 Join date : 20090915 Age : 31 Location : the infinite, frictionless plane of uniform density
 Subject: Re: [SOLVED] powers of 2 Fri Nov 20, 2009 9:11 am  
 Haha. Okay, sure :
Well the quotient between the greatest and the smallest is a power of 2. This quotient can't be a power of 2 greater than 8 or else the biggest of the two numbers would have more digits than the other. So the biggest of the two is either twice the other, four times the other, or eight times the other.
Moreover both numbers are congruent (mod 9) because every positive integer is congruent to the sum of its decimal digits (mod 9). So the difference between the two must be divisible by 9. But none of
2^(m+3)2^m = 2^m x 7,
2^(m+2)2^m = 2^m x 3,
2^(m+1)2^m = 2^m
are divisible by 9. 

 
peyman Euclid
Posts : 49 Join date : 20091106 Age : 94 Location : dense in the universe
 Subject: Re: [SOLVED] powers of 2 Fri Nov 20, 2009 2:09 pm  
 very good! (you're starting to solve problems like mohammad, saying everything is obvious!) 

 
Bruno Admin
Posts : 184 Join date : 20090915 Age : 31 Location : the infinite, frictionless plane of uniform density
 Subject: Re: [SOLVED] powers of 2 Fri Nov 20, 2009 2:17 pm  
  peyman wrote:
(you're starting to solve problems like mohammad, saying everything is obvious!)
Everything is obvious until proved otherwise, right? 

 
peyman Euclid
Posts : 49 Join date : 20091106 Age : 94 Location : dense in the universe
 Subject: Re: [SOLVED] powers of 2 Fri Nov 20, 2009 2:21 pm  
 

 
Mohammad Descartes
Posts : 100 Join date : 20091105 Age : 34 Location : Right behind you
 Subject: Re: [SOLVED] powers of 2 Fri Nov 20, 2009 2:46 pm  
 Bruno, you should start saying also "we can simply show". I used this sentence often in measure theory assignments and Prof Satncu writing me always "no wast of paper here" or "a bit of rush here" , and I also got slapped from the guy was training us math competition in high school for saying "it is obvious". 

 
Bruno Admin
Posts : 184 Join date : 20090915 Age : 31 Location : the infinite, frictionless plane of uniform density
 
 
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 Subject: Re: [SOLVED] powers of 2  
 

 
 [SOLVED] powers of 2  
