Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As usual, for a positive integer $n$, let $\sigma(n)$ denote the sum of all positive divisors of $n$ (including 1 and $n$).

What is the smallest ODD number such that $\sigma(n) \ge 3n$?

For comparison, the answers to some related questions are:

Smallest $n$ of any parity satisifying $\sigma(n) \ge 3n$ is $n=120=2^3\cdot3\cdot5$. We have $\sigma(120)=360$.

Smallest $n$ of any parity satisifying $\sigma(n) > 3n$ is $n=180=2^2\cdot3^2\cdot5$. We have $\sigma(180)=546$.

Smallest ODD $n$ satisifying $\sigma(n) \ge 2n$ is $n=945=3^3\cdot5\cdot7$. We have $\sigma(945)=1920$.

I suspect that the answer to my above question is $$ 1310112879075 = 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 $$ but I can't quite prove it.

I can prove that the answer must be divisible by each of the primes from 3 to 23, and then I can use trial and error to compare various candidates (e.g. compare multiplication by 29 with multiplication by $3^3$, and such "playing around").

share|improve this question
1  
I do not think that in general this kind of problem can be solved without some playing around. The fine structure of the order relations between powers of the primes is a mystery. –  André Nicolas Feb 21 '12 at 17:27
add comment

1 Answer

up vote 5 down vote accepted

Nope, the smallest is $$ n=1018976683725=3^3 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29, $$ which has $\frac{\sigma(n)}{n}=\frac{34283520}{11350339}\approx 3.02048423399513$.

Just for the record, I found this by using your observation that any candidate must be a multiple of $m:=3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23$. Your answer is $9765m$, and by running a for loop over all smaller (odd) multiples, I came upon the (unique) smaller answer of $n=9135m$.

share|improve this answer
1  
Thank you. That is clearly smaller than my answer, and also satisfies sigma(n) > 3n. A minor quibble: I proved that any candidate is divisible by all primes up to 23, not necessarily 29. However, even if we redefine m, the same trick of looping through odd multiples of m still verifies that we've found the smallest. –  idmercer Feb 21 '12 at 18:01
    
Oops, good catch. Edited for posterity. –  Cam McLeman Feb 21 '12 at 18:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.