Find how many integral solutions and there to the given condition for $x_1 , x_2 , x_3$ and $x_4$

$$x_1 \cdot x_2 \cdot x_3 \cdot x_4 = 210$$

I factored it to $2 \cdot 7 \cdot 5 \cdot 3$, Then how do I proceed?

  • 4
    $\begingroup$ Is the solution $x_1=1, x_2=70,x_3=3,x_4=1$ to be considered different from $x_1=3, x_2=1,x_3=1,x_4=70$? $\endgroup$ Jan 27, 2015 at 9:59

1 Answer 1


Since the prime factorization of $$210=2\cdot 3\cdot 5\cdot 7$$
What we have to basically do is find out the number of ways we distribute these factors among the four variables $x_1,x_2,x_3$ and $x_4$.
We can distribute each factor among the $4$ variables by ${4\choose 1}=4$ ways.
Hence, the answer is
Also considering negative numbers, we can multiply the result by
$$({4\choose 0}+{4\choose 2}+{4\choose 4})=8$$
This is because for negative numbers, only $2$ or all $4$ variables have to be negative to obtain positive product.
So, the final answer will be
$$256\times 8=2048$$

  • 1
    $\begingroup$ You need $\binom 40$ in there as well to capture the positive solutions! $\endgroup$ Jan 27, 2015 at 11:28
  • $\begingroup$ Oh... for some reason I thought that ${4\choose 0}=0$. Silly me. Thanks @MarkBennet $\endgroup$
    – AvZ
    Jan 27, 2015 at 11:30
  • 1
    $\begingroup$ Thanks !! Well I figured out I took 210= 3*5*2*7 and so x1x2x3x4 must be of the same form .. So we can write x1= 3^a1 * 5 ^b1 * 7^c1 * 2^d1 and so on for x2,x3 , and x4 so multiplying we want no of solution to equation , 3^(a1+a2..a4=1) * 5^(b1+b2..+b4=1) * 7^(c1+..+c4=1)* 2^(d1+..+d4=1) which can be easily found out via multinomial theorem.. Or dividers and balls method.. Giving us 4 for each so 256 and my logic for negative integers is same.. Giving ans 2048.. ! $\endgroup$ Feb 19, 2015 at 4:05
  • $\begingroup$ Although its basically the same thing.. $\endgroup$ Feb 19, 2015 at 4:09
  • $\begingroup$ I know it's late but you didn't include the cases in which one of the solution is $1$ and the other might be $\mathrm{(2\times 3, 5, 7)\text{ or }(2\times 5, 3, 7)\text{ or }(2\times 7, 3, 5)}$ and so on. $\endgroup$
    – Eyy boss
    Dec 15, 2020 at 16:31

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