Number of solutions of $100x+10y+z=5xyz$

Question

Find the number of solutions to the equation $$100x+10y+z=5xyz$$ where $x,y,z \in \mathbb{Z}$ and $(1 \le x \le 9)$, $(0 \le y \le 9)$, $(0 \le z \le 9)$.

I have found one solution using a brute-force computer program, but I've been wondering if it's possible to find the number of solutions without checking all possible triplets.

Well, $5,|,z$ so you just have to consider $z=0$ and $z=5$. $z=0$ is trivial, so you just need to consider $z=5$. — lulu, May 08 '23 at 16:32
@janq0 Welcome to Math SE. FYI, using an Approach0 search, there's the related Number theory - Find the number satisfying the condition.. However, for your particular question, I suggest first using lulu's comment before trying anything else suggested in that question or answers. — John Omielan, May 08 '23 at 16:43

score 2 · Answer 1 · answered May 08 '23 at 16:55

2

Clearly $z$ is a multiple of $5$ and $z=0$ does not yield a solution, so $z=5$. Then $$100x+10y+5=25xy.$$ A bit of algebra then shows that $$(5x-2)(5y-20)=25xy-100x-10y+40=45,$$ and then dividing by $5$ yields $$(5x-2)(y-4)=9.$$ It follows that $x=1$ and $y=7$.

answered May 08 '23 at 16:55

Servaes

63,261
7
75
163

1

Alternatively, remark that (with $z=5$) we have $2y+1\equiv 0\pmod 5$ so $y\in {2,7}$. – lulu May 08 '23 at 16:59
@lulu Really, I can not see any mathematical reason $z=5k$ implies $z=0\vee z=5$. Putting $z=5k$ doesn't obviously tell me $z=5$. What I am missing ? – lone student May 08 '23 at 17:10
2

@lonestudent The question specifies that $0\leq z\leq 9$. – Servaes May 08 '23 at 17:11
Then, nice answer . (+1) – lone student May 14 '23 at 01:24

Dan · Accepted Answer · 2023-05-08T22:11:21.350

If you do arithmetic modulo 5, then all the multiple-of-5 terms ($100x$, $10y$, and $5xyz$) cancel out and you're left with just $z = 0$ (mod 5). IOW, $z$ is a multiple of 5. There are two cases to deal with:

$$z = 0 \implies 100x+10y=0$$ $$z = 5 \implies 100x+10y+5=25xy$$

For the first case, $100x+10y=0$ is equivalent to $y = -10x$. So $y$ must be a multiple of 10, and the only in-range one is $y = 0$. But this gives us $x = 0$, which contradicts the given constraint $1 \le x \le 9$. So any solutions with $z = 0$ are ruled out.

For the $z = 5$ case, solving for $y$ in terms of $x$ gives:

$$y = \frac{20x+1}{5x-2}$$

Or equivalently,

$$y = 4 + \frac{9}{5x-2}$$

So in order for $y$ to be an integer, we must have $5x - 2$ be a divisor of 9. IOW, one of the following:

$5x - 2 = -9 \implies x = -1.4$
$5x - 2 = -3 \implies x = -0.2$
$5x - 2 = -1 \implies x = 0.2$
$5x - 2 = 1 \implies x = 0.6$
$5x - 2 = 3 \implies x = 1$
$5x - 2 = 9 \implies x = 2.2$

Only one of these results in an integer: $x = 1$, from which $y = \frac{21}{3} = 7$. Therefore, the only valid all-integer solution is:

$$\boxed{x = 1, y = 7, z = 5}$$

Note that $y$ is an integer and $$y=\frac{20x+1}{5x-2}=4+\frac{9}{5x-2},$$ so $5x-2$ divides $9$, so $x=1$. — Servaes, May 08 '23 at 17:13
@Servaes: Good point. There are only 6 possible divisors of 9 ($\pm{1, 3, 9}$), which saves some arithmetic compared to brute-forcing all 9 valid $x$'s, so I've edited my answer to use it. — Dan, May 08 '23 at 22:13
Given that $x\geq1$ it is immediate that the only candidate divisors are $3$ and $9$. — Servaes, May 09 '23 at 09:44

Number of solutions of $100x+10y+z=5xyz$

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