The question is easy.
My proof: If say some of the numbers is divisible by n then we are done. If no then there must be two numbers which must be congruent to the same number x(mod n). Let those two numbers be (2^i-1) and (2^j-1). Then subtract the smaller from the larger and factorize. WLOG let i>j. Then u get 2^j common outside and inside the bracket u get (2^(i-j)-1). Now since n is odd, gcd(n,2^j)=1. So n divides (2^(i-j)-1) and note that this quantity lies in the set and hence you are done :)

Sorry for not using Latex as I am in a hurry right now !

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TopNewestThe question is easy. My proof: If say some of the numbers is divisible by n then we are done. If no then there must be two numbers which must be congruent to the same number x(mod n). Let those two numbers be (2^i-1) and (2^j-1). Then subtract the smaller from the larger and factorize. WLOG let i>j. Then u get 2^j common outside and inside the bracket u get (2^(i-j)-1). Now since n is odd, gcd(n,2^j)=1. So n divides (2^(i-j)-1) and note that this quantity lies in the set and hence you are done :)

Sorry for not using Latex as I am in a hurry right now !

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If I can ask,are you a RMO participant?

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Seems like I have got a tough competitor from my state ;)

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Yes

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That is a nice solution and no problem!

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