Actual math problem re: Number theory regarding divisibility rules Here is the problem that supposedly has a solution and it is an extra credit problem for my 6th grade godchild. None of us can figure it out :(
Q: For the number ABC, each distinct letter represents a different digit. ABC, CAB, and BCA are all divisible by 6 and 9.
What are the values of A, B, and C
She's allowed to get help with this problem and I'm usually decent at word problems but I now have a headache...Please help!
Thank you!!!!!!!
 A: ABC, CAB, BCA divisible by 6 implies A, B, and C are all even numbers. Divisibilty by 9 implies A+B+C is divisible by 9. That's a start.
A: Here are some divisibility rules: 


*

*If a number is divisible by 9, then the sum of its digits is divisible by 9.

*If a number is divisible by 2, then its last digit is even.


In particular: $A+B+C$ is divisible by 9, and $A$, $B$, and $C$ are all even. Can you find three distinct $A,B,C$ that satisfy these properties?
A: Here's another way to solve the problem. If $ABC$, $CAB$, and $BCA$ are all divisible by 6, they're all even and end in an even digit. Therefore $A$, $B$, and $C$ are even. Then, $ABC$, $CAB$, and $BCA$ are even numbers divisible by 9, so they are multiples of 18. The smallest of the three digits must be 2 or 4, since you can't have distinct even digits all at least 6.
The three-digit multiples of 18 that begin with 2 or 4 can be computed by hand: 216, 234, 252, 270, 288, 414, 432, 450, 468, and 486. The only ones made up of three different even digits are 468 and 486, so the only possibilities for $A$, $B$, and $C$ are 4, 6, and 8 in some order. You can check that this solution works by hand. 
If three-digit numbers beginning with 0 are allowed, the possibility that the smallest of $A$, $B$, and $C$ is 0 can be ruled out, because the two-digit multiples of 18 all begin with odd digits.
A: ‘ABC’, when expanded in polynomial form, is $100A + 10B + C$.
Furthermore, $100A + 10B + C = (99A + A) + [9B + B] + C = 9(11A + B) + A + B + C$.
If ‘ABC’ is divisible by $9$, then $9(11A + B) + A + B + C = 9n$ for some integer $n$.
i.e. $A + B + C = 9[n - (11A + B)]$. This further implies $A + B + C$ is divisible by $9$.
If ‘ABC’ is divisible by $6$, then it is divisible by $2$ (and also by $3$).
This means ‘ABC’ is an even number. This further means that C is even.
Similarly, B and A are all even.
The whole question now boils down to
“find $3$ different even integers (each less than $10$) such that their sum is equal to a multiple of $9$.”
Because of the restrictions on A, B, C, the possible candidate(s)


*

*cannot be $9$ or $27$ because they are odd.

*cannot be over $30$ because A, B and C are all less than $10$.
Now the question becomes
“find 3 different even integers (each less than $10$) such that their sum is equal to $18$”. 
It is then not that difficult to guess what  the values of A, B and C are.
Note. The question is therefore suitable for kids excluding the analysis in the first paragraph. However, most likely, they have already been told the “secret” of testing whether a number is divisible by 9 or not.
A: A, B and C are all different even single digit numbers. 
A+B+C is divisible by 9.
Only answer is $4,6,8$
