How do you determine the total amount of positive factors of a number

**TravisH82** · October 6th 2009, 07:57 PM

Part of a homework assignment I'm stuck on from my discrete math class. I suspect this may belong in another forum, but I wasn't sure which one.

Part a.) Find the prime factorization of 7056. okay.. no problem
$7056 = 2 * 3528 = 2^2 * 1764 = 2^3 * 882 = 2^4 * 441 = 2^4 * 3 * 147 = 2^4 * 3^2 * 49 = 2^4 * 3^2 * 7^2$

Part b.) How many positive factors are there of 7056?
I'm stuck on this, I used an example of a previous question from the homework where I discovered that:
$48 = 2^4 * 3$ and that it has 10 positive factors. (namely 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48) and 10 negative identical factors.

I can't seem to derive the method to determine total factors based on the powers of each individual factor. I certainly don't want to write them all out. My best guess is 36, which i figured by adding each exponent, then adding each exponent multiplied to the other exponents, in turn, then adding all three exponents multiplied. (1+4+2+2+1)+(4*2+4*2+2*2)+(4*2*2)=10+20+16=36

something seems fishey about my methodology though..

Thanks for the help!

**Gamma** · October 6th 2009, 10:34 PM

just a counting argument. any factor of that number is going to share some or all of the prime factors.

Let $n=p_1^{e_1}\cdot... \cdot p_n^{e_1}$
Then any factor of n will have 0 factors of $p_1$ , 1 factor of $p_1$ , ... or $e_1$ factors of $p_1$ .

Similarly for each prime.

Thus the total number of possible combinations of factors will be
$(e_1+1)\cdot ... \cdot (e_n+1)$

So in your example you have $n=2^43^1$
so it has $(4+1)(1+1)=5\cdot2=10$ factors

**Soroban** · October 7th 2009, 06:03 AM

Hello, TravisH82!

Part (a) Find the prime factorization of 7056.

okay.. no problem . . . $7056 \:=\: 2^4\cdot3^2\cdot7^2$

Part (b) How many positive factors are there of 7056?

Let's construct a factor of 7056 . . .

We have 5 choices for twos: . $\begin{Bmatrix}\text{0 twos} \\ \text{1 two} \\ \text{2 twos} \\ \text{3 two's} \\ \text{4 two's} \end{Bmatrix}$

We have 3 choices for threes: . $\begin{Bmatrix}\text{0 threes} \\ \text{1 three} \\ \text{2 threes} \end{Bmatrix}$

We have 3 choices for sevens: . $\begin{Bmatrix}\text{0 sevens} \\ \text{1 seven} \\ \text{2 sevens} \end{Bmatrix}$

Hence, there are: . $5\cdot3\cdot3 \:=\:45$ choices for a factor of 7056.

. . Therefore, 7056 has $\boxed{45}$ factors.
. . . This includes 1 and 7056 itself.

Now you can see why Gamma's formula works.

Find the prime factorization: . $7056 \;=\;2^4\cdot3^2\cdot7^2$

Add 1 to each exponent and multiply:

. . $(4+1)(2+1)(2+1) \;=\;5\cdot3\cdot3 \;=\;45$ factors . . . see?

**Gamma** · October 7th 2009, 07:57 AM

Exactly. Wow very nice looking solution Soroban, exactly what I was going for, but I suck at TeX, lol. Thanks for taking the time to clean it up.

**TravisH82** · October 13th 2009, 06:21 PM

Ahh.. I see where I was going wrong.. i was neglecting to consider 2^0 as a choice. Thank you all for your help.. this makes it much clearer to me!

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