The five digit number $xy26z$ (each of the letters represents a digit) is divisible by $8$, $9$ and $11$. What is the value of $x$?

Question

The only thing I came up with was using the divisibility criteria, so: $$8 + z \equiv 0 \pmod{8} \implies z \equiv 0 \pmod 8$$

$$x + y + 8 + z \equiv 0 \pmod{9}$$

$$x - y - 4 + z \equiv 0 \pmod{11}$$

But that's as far as I've got. I'd appreciate some help.

Answer 1

$z \equiv 0 \pmod 8$ is not correct because $260$ is not divisible by $8$. Once you fix that, you will have three equations in three unknowns. You should get $z$ directly because it has to be a single digit. That leaves two equations in two unknowns. The $\pmod 9$ one can only sum to $18$ or $27$. The $\pmod {11}$ can only sum to ??? given that $x,y$ are single digits.

Answer 2

The divisibility test for $8$ is whether the last three digits are divisible by $8$. So you must have $26z$ a multiple of $8$. Hence $z=4.$

By the test for $11$ we have

$$x-y+2-6+4 \equiv 0 \pmod{11}$$

or

$$x-y \equiv 0 \pmod{11}.$$

So we must have $x=y$.

By the test for $9$ we have

$$x+x+2+6+4 \equiv 0 \pmod{9}$$

or

$$2x \equiv 6 \pmod{9}.$$

So $2x =6$ or $2x=15.$ So $x=3=y$. Final answer $33264.$

Answer 3

$xy26z=z+6+2+y+x+6\times 9+2\times 99+y(10^3-1)+x(10^4-1)$

So for divisibility by 9 we must have following relation:

$z+x+y+8=9 k_1$

$(10^4-1)$ and $(100-1)$ are divisible by 11 so we must have:

$3000+60+2+x+z=11 k_2$

For $k_2=279$ we get:

$x+z=7$

Note: we may consider the remainder of 3062 divided by 11:

$3062≡4 \mod 11$ ⇒ $4+x+z=11k_2$, with $k_2=1$ we get $x+z=11-4=7$

Since z must be even the possible values of x and z are:

$(x, z)=(3, 4), (5, 2), (1, 6), (6, 0)$

So first relation reduces to:

$y+8+7=9k_1$

Clearly we must have $k_1>1$, let $k_1=2$ we get:

$y=18-15=3$

By subtracting two relation we get:

$x+z-(x+y)=z-y=7-6=1$ ⇒ $z=4$

Therefor in $(x, z)$ couples only $(x, z)=(3, 4)$ is acceptable, hence the number is $33264$.