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### Finding Numbers with a Certain Number of Factors

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Date: 3/12/96 at 16:18:46
From: Patrick Krause
Subject: Factoring Question

Dr. Math:

I am stuck on the following question:

Given that twelve is the least postive integer with six
different positive factors (1,2,3,4,6,12).  What is the
least positive integer with exactly twenty-four positive
factors?

I know by writing a factoring program that the answer is
360 (factors:

1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180,360)

but I cannot make the connection between the factors of 12
and the way to find a number with 24 positive factors.

I have tried prime factorization, I have tried everything that
I can to make this connection, but I have cannot make the
connection.  I do not see what knowing the factors of 12 have
to do with the factors of 360.

Can you give me some help with this?

Thank you.
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Date: 5/30/96 at 11:38:32
From: Doctor Alain
Subject: Re: Factoring Question

Let P1, P2, ... , Pn be the prime factors of a given number X, and
let E1, E2, ..., En be the greatest exponents of the respective
prime factors such that their powers are still factors of X. That
is to say that

X = P1^E1 * P2^E2 * ... * Pn^En.

Then all the positive factors F of X are in the form

F = P1^e1 * P2^e2 * ... * Pn^en,

where e1, e2, ... , en are all integer numbers greater or equal to
zero and less than or equal to their respective Ei (e1 less than
or equal to E1, e2 less than or equal to E2, ... , en  less than
or equal to En). Any set of integers e1, e2, ... , en satisfying
these conditions will produce a factor of X. The set of numbers
satisfying these conditions is one-to-one onto with the set of
positive factors of X. So the number N of positive factors of X
is given by

N = (E1+1) * (E2+1) * ... * (En+1).

[There are E1+1 ways to chose e1, E2+1 ways to chose e2, ...,
En+1 ways to chose en]

Now let's go back to your problem. You want a number with 24
positive factors, 24 is 3*8 = 3*4*2 = 3*2*2*2 so the numbers E1,
E2, ... , En will be either

(#1) E1=2 (E1+1=3), E2=7 (E2+1=8)       [24 = 3*8]
or
(#2) E1=2, E2=3, E3=1                   [24 = 3*4*2]
or
(#3) E1=2, E2=1, E3=1, E4=1.            [24 = 3*2*2*2]

These are the only ways you can get N = (E1+1) * (E2+1) * ... *
(En+1)=24.

So all integers X with 24 positive factors can be written in one
of the following forms

(#1) X = P1^2 * P2^7

(#2) X = P1^2 * P2^3 * P3

(#3) X = P1^2 * P2 * P3 * P4

The smallest number of form #1 is obviously 3^2 * 2^7 = 1152.
The smallest number of form #2 is obviously 3^2 * 2^3 * 5 = 360.
The smallest number of form #3 is obviously 2^2 * 3 * 5 * 7 = 420.

-Doctor Alain,  The Math Forum

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Associated Topics:
High School Discrete Mathematics

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