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Guest Message by DevFuse
 

Barcallica

Member Since 27 Apr 2012
Offline Last Active Feb 27 2015 09:04 AM
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Posts I've Made

In Topic: form: 36^k - 5^m

18 February 2015 - 03:05 AM

 

k=1 m=2 smallest 11.


Now to prove your answer is the smallest.

 

 

Last digit of result always is 1. We should prove it cannot be 1.

36k = 5m +1, left side is dividable by 4. Right side is not.


In Topic: form: 36^k - 5^m

17 February 2015 - 03:43 AM

it does not need to be natural, only k and M have to.

 

k=1 m=-infinite (though -infinite is not real number)


In Topic: form: 36^k - 5^m

16 February 2015 - 08:28 AM

k=1 m=2 smallest 11.


In Topic: form: 36^k - 5^m

16 February 2015 - 08:08 AM

 

Among all the numbers representable as 36k - 5m (k and m are natural numbers) find the smallest. 
Prove that it is really the smallest.

 

 

You mean smallest natural?


In Topic: The n-gon eats out

03 February 2015 - 04:08 AM

Spoiler for proof by contradiction

 

I was following the topic to see the solution to this seemingly simple but very interesting problem. 

And very nice solution by plasmid. Even school kids could understand without any trouble.  :thumbsup: