What are the possible integer values for $x,y,z$ if $792|\overline{5x36y98z2}$?

Question

So the question is Let $x,y, z$ Be digits so that the 9-digit number 5$x$36$y$98$z$2 is divisible by 792.

1) Without calculating $x$ and $y$, explain why z = 3 or 7.

2) WIthout calculating $x$, show that y = 0 or 9.

3) FInd all combinations of $x, y, z$

Answer 1

$792$ is also divisible by $9$ and $11$ as well.

Using the divisibility rule for $11$, $5-x+3-6+y-9+8-z+2$ or $-x+y-z+3$ must be divisible by $11$. What multiples of $11$ can this expression be?

In addition, using the divisibility rule for $9$, $5+x+3+6+y+9+8+z+2$ must be divisible by $9$. Therefore $x+y+z+33$ must be a multiple of $3$, which implies that $x+y+z$ must be a multiple of $3$ as well.

Then you can consider what happens when $z$ is $3$ or $7$.

Answer 2

by a simple program we can find all values for $x$, $y$, and $z$.

# n = 5x36y98z2
for x in range(10):
    for y in range(10):
        for z in range(10):
            n = 2 + 10*z + 8*10**2 + 9*10**3 + y*10**4 + 6*10**5 + 3*10**6 + x*10**7 + 5*10**8
            if n % 792 == 0:
                print x, y, z

So there are only three solution: $$x = 0, \quad y = 0,\quad z = 3$$ and $$x = 5, \quad y = 9,\quad z = 7$$ and $$x = 9, \quad y = 9,\quad z = 3$$

Answer 3

$2^{\large 3}\mid 10^{\large 3}$ so $\ 8\mid 5x36y98z2\iff 8\mid 800\!+\!10z\!+\!2$ $\iff 4\mid 5z\!+\!1\iff 4\mid z\!+\!1\iff \color{#c00}{z = 3,7}$

$\!\bmod 99\!:\,\ 100\equiv 1\,$ so $\ 0\equiv n\equiv 5\! +\!x3\! +\! 6y\! -1\! +\! z2 \equiv (x\!+\!6\!+\!z)10 + 9\!+\!y =: \color{#0a0}{\bar n\, <\, 3\cdot 99}\,$

so either $\,\color{#0a0}{\bar n = 99},\ $ so $\,y=0,\ x+\color{#c00}z = 3,\ \ \ $ so $\ \color{#c00}{z=3},\ x = 0$

or else: $\ \color{#0a0}{\bar n= 198},\ $ so $\ y = 9,\ x+\color{#c00}z = 12,\ $ so $\ \color{#c00}{z = 3},\ x = 9,\ $ or $\ \color{#c00}{z = 7},\ x = 5$