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Prove that among any 12 consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer n are all positive integers other than 1 and n which divide n. For example, the proper divisors of 14 are 2 and 7)

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Hint: Among any $12$ consecutive positive integers, there is one that is a multiple of $12$.

Can you show that $12n$ is smaller than the sum of its divisors for any positive integer $n$?

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  • $\begingroup$ $d(12)=2+3+4+6=15>12$... Ooops, I read the question "opposite" (which is why I had a hard time proving it) $\endgroup$ – barak manos Jun 28 '16 at 7:07
  • $\begingroup$ ^Yes, $12$ is smaller than the sum of its proper divisors. EDIT: Yeah, I initially read the question the other way as well, which caused me to take a long time with it. $\endgroup$ – JimmyK4542 Jun 28 '16 at 7:09

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