Factoring Strategies

Date: 11/01/2002 at 09:02:37
From: Verna Snedacar
Subject: Factoring strategies

Hi Dr. Math, 

I am returning to school and I am having a very hard time in a math 
for elememtary teachers class. The objective is to teach for 
understanding.

My assignment is to explore strategies that help you find a number 
having 5 factors and a number having 13 factors and to explain my 
thinking and strategies. 

So far I think I have found that the number has to be a square number 
because square numbers have an odd number of factors. For example, 
16 = 1, 2, 4, 8, and 16, but I need to know why. Is there a way I can 
know a number has 13 factors without wrecking my brain factoring all 
of the square numbers? 

Please don't use x = y z and so on; I do not understand what all that 
means.  I also don't know what this symbol means (^) as I explore your 
site.

I really need some help. Thanks.

Date: 11/01/2002 at 13:19:49
From: Doctor Peterson
Subject: Re: Factoring strategies

Hi, Verna. 

It would be a great idea to become at least somewhat comfortable with 
the idea of variables (x, y, z, and so on), because they are an 
extremely useful part of the language of mathematics. As for a^b, 
you can find it explained in our FAQ:

   Typing math
   http://mathforum.org/dr.math/faq/faq.typing.math.html 

We use it to represent exponents without having to take up two lines; 
3^2 means 3 squared.

It's hard to discuss this problem without using variables; that's sort 
of like telling certain people to talk without moving their hands! But 
I'll try. That means I'll mostly have to use examples.

Let's look at the factors of a square number, first; say, 6^2 = 36:

    1*36
    2*18
    3*12
    4*9
    6*6
    9*4
    12*3
    18*2
    36*1

Here I listed not just the factors, but the factor pairs. That is, for 
each factor, there is another number you multiply it by to get 36, and 
I included it in my list. We have 9 factors, and 9 factor pairs.

Notice that every factor appears twice in the list, once as the first 
member of a pair, and once as the second. That means that the rows can 
be paired off - except for the 6*6 row:

    1*36 --------+
    2*18 ------+ |
    3*12 ----+ | |
    4*9  --+ | | |
    6*6  o | | | |
    9*4  --+ | | |
    12*3 ----+ | |
    18*2 ------+ |
    36*1 --------+

The only exception is the 6*6 row, because there both occurrences of 6 
are in the same row.

Now, when you can pair everything up, you have an even number; when 
you can't you have an odd number. In this case, we have an odd number 
of rows, namely 9 factors. If the number were not a square, we would 
have an even number of rows, because there would be no special row 
like 6*6. For example, here are the factor pairs for 35:

    1*35 ---+
    5*7  -+ |
    7*5  -+ |
    35*1 ---+

Here every factor pairs up with another in a different row, and there 
is an even number of them.

Does that demonstrate why you need a square in order to have an odd 
number of factors?

Now things will get a little more complicated, because I'm going to 
show how you can find the number of factors without listing them. 
Let's work with 36 again. First we'll find the prime factorization:

    36 = 2*2*3*3

Now, when we make a factor pair, what we are really doing is just 
splitting these prime factors up between two numbers. For example, 
the factor pair 3*12 is

    36 = 3 * 2*2*3

where I took 3 for the first factor, and all the rest for the second. 
Do you see why this has to be true? When you multiply two numbers 
together, the prime factors of the product are the prime factors of 
the factors, combined; we're just pulling them apart.

Now, how many ways can you split up the prime factors of a number? 
Let's make a table, listing how many of each prime factor we use for 
the first factor:

     number   number   product of
     of 2's   of 3's   2's and 3's
    --------+--------+------------
        0   |   0    | 2^0*3^0 = 1
        0   |   1    | 2^0*3^1 = 3
        0   |   2    | 2^0*3^2 = 9
        1   |   0    | 2^1*3^0 = 2
        1   |   1    | 2^1*3^1 = 6
        1   |   2    | 2^1*3^2 = 18
        2   |   0    | 2^2*3^0 = 4
        2   |   1    | 2^2*3^1 = 12
        2   |   2    | 2^2*3^2 = 36

I can choose anything up to the maximum number of 2's in 36, namely 0, 
1, or 2 of them; and the same with the 3's. Altogether, this gives me 
3 times 3, or 9, ways to choose the prime factors I want to use; and 
that gives me 9 factors of 36.

Just in case you're not fully comfortable with powers (especially when 
the exponent is 0), here's another way to look at the same thing:

                          product of
       2's     3's        2's and 3's
    --------+--------+-------------------
            |        | 1             = 1
            |   3    | 1 *       3   = 3
            |  3*3   | 1 *       3*3 = 9
        2   |        | 1 * 2         = 2
        2   |   3    | 1 * 2   * 3   = 6
        2   |  3*3   | 1 * 2   * 3*3 = 18
       2*2  |        | 1 * 2*2       = 4
       2*2  |   3    | 1 * 2*2 * 3   = 12
       2*2  |  3*3   | 1 * 2*2 * 3*3 = 36

I'll leave it to you to generalize this idea, and try it on some other 
numbers. The idea is that you count the number of each prime factor in 
your number; the number of factors will be one more than the number of 
each prime factor, all multiplied together. In this case, we had two 
2's and two 3's, so (2+1)*(2+1) = 3*3 gives 9 factors. For 35, we have 
one 5 and one 7, so the number of factors is (1+1)*(1+1) = 2*2 = 4, as 
we saw.

Now you need to apply this to your problem. If a number has 13 
factors, what does that tell you about its prime factors? Can you find 
a number like this? Can you find more of them?

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/