Integral Numbers Question

Question

How many integer solutions are there to xyz=120? This is what I have tried so far: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=609981 see the first solution in this thread. Any help would be appreciated. Thank you.

Answer 1

Write that $120 = 2^3\times 5^1\times 3^1$. A decomposition $120 = abc$ implies

• $a = 2^{a_1}3^{a_2}5^{a_3}$
• $b = 2^{b_1}3^{b_2}5^{b_3}$
• $c = 2^{c_1}3^{c_2}5^{c_3}$ with $a_1 + b_1 + c_1 = 3$, $a_2 + b_2 + c_2 = 1$, $a_3 + b_3 + c_3 = 1$.

As all choices are independent, you get

• 3 choices for the values of $(a_2, b_2, c_2)$ and $(a_3, b_3, c_3)$ (which one is $\neq 0$)
• 10 choices for the value of $(a_1, b_1, c_1)$:

300; 210; 201; 120; 111; 102; 030; 021; 012; 003

hence a total of $3\times 3\times 10 = 90 $ choices.

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if counting negative solutions as well, you should consider for each non negative solution $a,b,c$ to the problem the solutions $a, b, c$; $-a, -b, c$; $-a, b, -c$; $a, -b, -c$; so just multiply by 4 the total.