11.1. What three digits should be appended to the right of the number 579 so that the resulting six-digit number is divisible by 5, 7, and 9? Find all possible solutions.
Answer. $(6,0,0),(2,8,5)$ or $(9,1,5)$. The obtained number will be, respectively, 579600, 579285, or 579915.

Solution. Let the digits to be appended be denoted as $x, y, z$ in order from left to right. The resulting 6-digit number will be $\overline{579 x y z}$. By the divisibility rule for 5, we have $z=0$ or $z=5$. By the divisibility rule for 9, in the first case $x+y$ gives a remainder of 6 when divided by 9, and in the second case, a remainder of 1.

In the first case, $x+y$ is 6 or 15. The remainder from dividing the number $\overline{579 x y 0}$ by 7 is the same as the remainder from dividing the number $2+2x+3y$ by 7. Therefore, either $2+2x+3y=2+2x+3(6-x)=20-x$ is divisible by 7, leading to $x=6, y=0$, or $2+2x+3y=2+2x+3(15-x)=47-x$ is divisible by 7, leading to $x=5, y=10$, which is impossible.

In the second case, $x+y$ is 1 or 10. The remainder from dividing the number $\overline{579 x y 5}$ by 7 is the same as the remainder from dividing the number $2x+3y$ by 7. Therefore, either $2x+3y=2x+3(1-x)=3-x$ is divisible by 7, leading to $x=3, y=-2$, which is impossible, or $2x+3y=2x+3(10-x)=30-x$ is divisible by 7, leading to $x=2, y=8$, or $x=9, y=1$.

Evaluation. Just the answer with verification: 1 point. Loss of each solution minus 2 points. Unjustified statements like “$x+y$ is 6 or 15” minus 1 point.