Numbers with 12 Factors....

Date: 11/15/96 at 02:59:25
From: Shane Conrad
Subject: Numbers with 12 Factors....

I have to find two numbers which have exactly 12 factors, including 
1, the number itself, 5 and 7. I found one such number, 140 (140 has 
factors of 1,2,4,5,7,10,14,20,28,35,70,140) but I had to do a lot of 
guessing.  Is there another way to figure problems like this out 
without trying a lot of numbers? 

Date: 04/01/97 at 19:39:52
From: Doctor Yiu
Subject: Re: Numbers with 12 Factors....

Dear Shane,

Suppose you factor a number into a product of prime powers:

   the first prime appearing  A times,
   the second prime appearing  B  times,
   ....
   the last prime appearing  Z  times.

Then that number has exactly 

   (1 + A) times (1 + B) times ... all the way up to (1+Z)

divisors, including 1 and itself.  This is a theorem in a subject
called number theory.  If you want to know why it holds, write back to 
us!

Since your number is divisible by 5 and 7,  you can take the 
first prime to be 5 and the second to be 7, and ask whether it can 
have other primes or not.  Note that 1+A and 1+B cannot be smaller 
than 2. Since there are altogether 12 divisors, these (1+A), (1+B), 
and others if any, should multiply to 12.

Now the only ways to write 12 as a product of numbers with at 
least two of them not smaller than 2 are

(i)  2  times  6
(ii) 3  times  4
(iii)  2  times  2 times 3

In (i) and (ii), the number has only 5 and 7 for its prime divisors. 
It must be

     5^1 times  7^5  =  84035,
or   5^5 times  7^1  =  21875,
or   5^2 times  7^3  =   8575,
or   5^3 times  7^2  =   6125.

In (iii), the number has one more PRIME divisor. Let it be  p.
This number is 

either  5 times 7 times p^2 = 35p^2,
or      5 times p times 7^2 = 245p,
or      p times 7 times 5^2 = 175p.

Your example corresponds to the first of these, with  p = 2.

I hope this helps.  If you are confused about anything, please do 
write back! Good luck.

-Doctors Yiu and Sydney,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/