Possible conjecture relating to factors and powers of 12

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I have read in the past that having a base 12 number system is superior to a decimal system because of the number of factors that 12 has. For instance we essentially use base 12 for time, it's better to have a 60 minute hour than a 100 minute hour as there are more 'convenient chunks' or factors of 1 hour with 60 minutes.

This led to me having a thought recently, are there any multiples of 12 for which there are smaller numbers with more factors? It looked briefly promising but it turns out there are lots of exceptions, for instance 132 (11*12) only has 10 factors, 126 has 12 factors. A few of the preceding multiples of 12 also have more factors than 132.

I then decided to try it with integer powers of 12. So far I haven't found any exceptions, all the integer powers of 12 appear to have lots of factors:

12: 6 factors
144: 15 factors
1728: 28 factors
20736: 45 factors

I can't find anywhere that will let me work out the number of factors for higher powers and don't have the programming skill to do so myself. (To get the value for the 2 larger values above I used http://www.mathsisfun.com/numbers/factors-all-tool.html" to get the factors and then pasted them into a blank word doc to quickly add them up using word count).

A neat way to pose the conjecture in words would be: There are no numbers smaller than a power of 12, x, with more factors than x.

Can anyone quickly disprove this? I'm not sure if it's even interesting at all but I got curious.

Edited for accuracy.
 
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Answers and Replies

  • #2
mathman
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11*12=132. 10 factors.
 
  • #3
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Whoops, it still has fewer factors than the previous 6 multiples of 12 (5*12 to 10*12). 132 also has fewer factors than 126 which has 12 factors. So I am still correct in saying it doesn't work for multiples.
 
  • #4
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  • #5
AlephZero
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The way to count the number of factors is:

Factorize the number completely, as 2^a 3^b 5^c ....
The number of factors is then (a+1)(b+1)(c+1) ....

e.g. 60 = 2^2.3.5
The number of factors = (2+1)(1+1)(1+1) = 12

For powers of 12, this gives
12 = 2^2.3 Number of factors 3x2 = 6
12^2 = 2^4.3^2. Number of factors 5 x 3 = 15
...
12^1000 = 2^2000.3^1000. Number of factors = 2001 x 1001 = 2003001

You can use this "in reverse" to find small numbers that have a given number of factors. For example if you want a number with 20 factors, then
20 = 5x4 = 5x2x2
So a number with 20 factos must be either p^4q^3 or p^4qr, where p q and r are different primes.

So two small numbers with 20 factors are 2^4.3^3 = 432 and 2^4.3.5 = 240.
 

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