Is there any way to prove this? rather than just plugging in numbers?
It's related to Mersenne Primes for anyone interested. I only wanna know the proof to the above statement.
Thank you.
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4Please specify what is $\sigma$ and what are your attempts to solve the problem. – the_candyman Mar 29 '15 at 11:46
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1Are you trying to write $\sigma(2^{k-1})=2^k-1$? – Gerry Myerson Mar 29 '15 at 11:47
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Garry Myerson, Yes, sorry, when I copy/pasted, it came out differently. But yes that is what I meant to write. – amin Mar 29 '15 at 11:54
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1And did you mean to write "Gerry", when you wrote "Garry"? – Gerry Myerson Mar 30 '15 at 00:54
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Hahah, yeah I did. :) – amin Mar 30 '15 at 16:27
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Hint: Every divisor of $2^{k-1}$ is of the form $2^i$ where $0\le i\le k-1$. Sum of these divisors forms a geometric series $$1+2+\dots +2^{k-1}=2^k -1$$
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