Divisible by 3 Riddle

Since $a,b,c$ are divisible by $3$ , let $a=3k_1, b=3k_2, c=3k_3$ , where $k_1,k_2,k_3$ are arbitrary integers and $k_2 \neq 0$ . Then, $\dfrac{a}{b} + c = \dfrac{k_1}{k_2} + 3k_3$ For above to be divisible by 3, this implies $\frac{k_1}{k_2}$ to be divisible by 3. Thus, $\begin{array}{rrl} &\dfrac{k_1}{k_2} &= 3n\\ &k_1 &= 3nk_2\\ \Longrightarrow& k_1 &= b\\ \Longrightarrow& a &= 3nb = 9nk_2\\ \end{array}$ where $n$ is an integer. This implies that $\boxed{a \text{ is divisible by 9}}$ .

Given the information in the problem we can write

$a=3\alpha\\b=3\beta\\c=3\gamma\\\frac{a}{b}+c=3\delta$

Solving the last of these for a and then substituting for b and c gives

$a=3\delta b-cb\\=9\delta \beta-9\gamma \beta\\=9(\delta \beta-\gamma \beta)$

and so a is a multiple of 9.

To complete the solution we need to show that it is NOT necessary for b and c to be multiples of 9. A single counterexample is enough. Taking

$a=9\\b=3\\c=3$

gives $\frac{a}{b}+c =6$

a/b + c = 3^n / 3 + 3n --> the whole equation won't be divisible by 3 if a/b = non-multiples of 3. Therefore, a/b must be larger or equal to multiples of 3, in this case, we can make it 3. It will still work for larger multiples of 3, which still proves that its divisible by 3^2: 3^n / 3 >= 3 n >= 2

So the power of (a) must be equal or larger than 2, which 3^2 = 9 --> divisible by 9