-3

I was wondering if there is any fast way to do the following problem:

Find the number of ordered triples $(x, y, z)$ to $$(x-2020)(2y-2021)(3z-2022)=9$$ where $x$, $y$, and $z$ are integers.

Remember: $x$, $y$, and $z$ can be negative!

random guy 2000
  • 139
  • 1
    That doesn't have any solutions, I think that third factor should be $3z-2022$. Assuming that, you can use the fact that the prime factorization of $9$ is $3^2$, so if $abc = 9$, where $a,b,c \in \mathbb{Z}$, then at least one of $a$, $b$, or $c$ must be $1$ or $-1$. Set each factor to $\pm 1$, and you've only got a handful of possible solutions. – Amaan M Jul 18 '21 at 18:47
  • That's what I meant, and thanks! – random guy 2000 Jul 18 '21 at 19:50

2 Answers2

0

Construct table

$x - 2020 = 3, -3 \implies x = 2023, 2017$
$x - 2020 = 1, -1 \implies x = 2021, 2019$

$2y - 2021 = 3, -3 \implies y = 1012, 1009$
$2y - 2021 = 1, -1 \implies y = 1011, 1010$

$z - 674 = 3, -3 \implies z = 677, 671$
$z - 674 = 1, -1 \implies z = 675, 673$

By considering factorizations of $3$, it becomes easy to find triplets of $(x,y,z)$, those being: $(2023,1011,675), (2023,1010,673), (2017, 1010, 675), (2017, 1011, 673)$
$(2021, 1012, 675), (2019, 1012, 673), (2021, 1009, 673), (2019, 1009, 675)$
$(2021, 1011, 677), (2019, 1010, 677), (2021, 1010, 671), (2019, 1011, 671)$

egglog
  • 1,674
-1

$(x-2020)(2y-2021)(3z-2022)=9$ can be simplified to $(x-2020)(2y-2021)(z-674)=3$, now you have to consider the factorizations of $3$:

$1\cdot 1 \cdot 3$

$1\cdot (-1) \cdot (-3)$

$(-1)\cdot 1 \cdot (-3)$

$(-1)\cdot (-1) \cdot 3$

$1\cdot 3 \cdot 1$

$1\cdot (-3) \cdot (-1)$

$(-1)\cdot 3 \cdot (-1)$

$(-1)\cdot (-3) \cdot 1$

$3\cdot 1 \cdot 1$

$3\cdot (-1) \cdot (-1)$

$(-3)\cdot 1 \cdot (-1)$

$(-3)\cdot (-1) \cdot 1$

that gives you $12$ solutions.

jjagmath
  • 18,214