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### 24 Factors

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Date: 12/08/96 at 05:39:50
From: Jim Fletcher
Subject: 24 different positive integers?

Dear Dr. Math,

impossible to prove solidly!

12 is the smallest positive integer with 6 different positive factors.
The factors are 1, 2, 3, 4, 6, and 12.  What is the least positive
integer with exactly 24 different positive factors?

Thank you very much.

Jim Fletcher
```

```
Date: 12/11/96 at 11:05:46
From: Doctor Rob
Subject: Re: Least positive integer with 24 different positive
factors?

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 integers greater than or equal to zero
and less than or equal to their respective Ei (e1 is less than or
equal to E1, e2 is less than or equal to E2, ... , en is 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 exactly 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:

24 = 12*2 = 8*3 = 6*4 = 6*2*2 = 4*3*2 = 3*2*2*2

(Factors of 1 can be ignored, since then Ei = 0, and the prime P1
would not divide the number.)  So the numbers E1, E2, ... , En will be
either:

(No.1) E1=23 (E1+1=24)                    [24 = 24]
or
(No.2) E1=11 (E1+1=12), E2=1 (E2+1=2)     [24 = 12*2]
or
(No.3) E1=7 (E1+1=8), E2=2 (E2+1=3)       [24 = 8*3]
or
(No.4) E1=5 (E1+1=6), E2=3 (E2+1=4)       [24 = 6*4]
or
(No.5) E1=5, E2=1, E3=1                   [24 = 6*2*2]
or
(No.6) E1=3, E2=2, E3=1                   [24 = 4*3*2]
or
(No.7) 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:

(No.1) X = P1^23

(No.2) X = P1^11 * P2

(No.3) X = P1^7 * P2^2

(No.4) X = P1^5 * P2^3

(No.5) X = P1^5 * P2 * P3

(No.6) X = P1^3 * P2^2 * P3

(No.7) X = P1^2 * P2 * P3 * P4

The smallest number of form No.1 is obviously 2^23 = 8388608.  I leave
it to you to figure out the smallest number of each of the other

-Doctor Rob,  The Math Forum (with a tip of the hat to Doctor Alain!)
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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