# Number of solutions $(x_1)(x_2)(x_3)(x_4) = 2016$

Having some trouble wrapping my head around this one:

find the number of solutions to the equation $(x_1)(x_2)(x_3)(x_4) = 2016$, where $(x_i)$s are integers that are not necessarily positive.

Here's what I've been thinking so far:

Let's bring $2016$ down to its prime factorized form, such that we have $2016 = 2^5\times3^2\times7^1$.

Because the solution may include negative integers, note that a pair or 4 integers will form positive numbers, therefore we have $C(4,2)$ OR $C(4,4)$ ways of doing this.

Let's also note that the solution may include 1s in $3, 2, 1$, or $0$ positions for $(x_i)$, such that we have $C(4,3), C(4,2), C(4,1)$, or $C(4,0)$.

Now I'm thinking that we should look at each particular element of the prime factorization, such that we have elements $2, 2, 2, 2, 3, 3, 3, 7$ that are not entirely distinct. We should now think of the number of ways we can fill in $8$ elements into each of these $1, 2, 3$, or $4$ positions depending on the number of 1s we have.

And this is where I am having a little bit of trouble thinking through the solution. I'm thinking I could check the number of distinct integers that I could form with $2^5\times3^2\times7^1$ and then go from there? But I'm having trouble thinking through this methodology as well.

Thoughts?

I will assume that you want the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ such that $x_1x_2x_3x_4=2016$.

First let us find the number of ordered quadruples of positive integers.

We want to distribute the five $2$'s between $x_1,x_2,x_3,x_4$.

By Stars and Bars there are $\binom{5+4-1}{4-1}$ ways to do this.

For each such choice we need to distribute the two $3$'s, and then the one $7$. I will assume you know now how to complete this part. Call the resulting number $N$.

To take account of negative possibilities, we could have all negative, giving another $N$ choices, or two positive and two negative, giving another $\binom{4}{2}N$ choices.

• So the final answer is $(1 + 1 + \binom{4}{2})N$ choices? – Eli Rose Jan 27 '16 at 23:54
• Yes. I am leaving the rest of the calculation of $N$ to you. If there is a problem, please leave a message. – André Nicolas Jan 27 '16 at 23:56
• Oh, I'm not the OP. Just wanted to make sure I understood :) – Eli Rose Jan 27 '16 at 23:58
• Apologies, I assumed, without checking, that you were. – André Nicolas Jan 28 '16 at 0:00
• Thank you for providing some intuition André. The link to Stars and Bars helped a lot. – Leeho Lim Jan 28 '16 at 5:01

The solution is hidden in 2^5*3^2*7 We have to select 4 factor whose multiplication is 2016

So for selecting 4 factor there are C(8,2)waysways

&you have to also include negative no Cases

• This might work better as a comment, as it seems to need some improvements to get the actual answer. See if learning MathJax and $\LaTeX$ to post mathematical expressions appeals to you. – hardmath Jan 28 '16 at 2:53