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)
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
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 $n$?
answered Jun 28, 2016 at 7:07
-
$\begingroup$ $d(12)=2+3+4+6=15>12$... Ooops, I read the question "opposite" (which is why I had a hard time proving it) $\endgroup$ Jun 28, 2016 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. $\endgroup$ Jun 28, 2016 at 7:09
-
$\begingroup$ In fact, among any 6 consecutive integers greater than 6 there is one such number, because $6n$ is strictly abundant for any $n$ strictly greater than 1. $\endgroup$ Jul 22, 2022 at 12:33
-
$\begingroup$ @barakmanos Since every 6th number, starting from 12, is abundant, there is a limit to the number of consecutive deficient number you can get. But if you reverse the inequality, there is no limit to the number of consecutive abundant numbers you can have. This is proved in comments by Jianing Song in A094268. $\endgroup$ Jul 22, 2022 at 12:55