# Divisibility ?

Jun 11, 2016

${n}_{1} = 12132$ and ${n}_{2} = 12636$

#### Explanation:

A number $n$ is divisible by $4$ when the number formed by its two right digits is divisible by $4$ and the same number $n$ is divisible by $9$ when the sum of its digits is divisible by $9$.

We have

${c}_{1} \to 3 \times 10 + \diamond = 4 \times {k}_{1}$
${c}_{2} \to 1 + 2 + \square + 3 + \diamond = 9 \times {k}_{2}$

here $\left\{\square , \diamond\right\} \in \left\{0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9\right\}$

From ${c}_{1} \to \diamond \in \left\{2 , 6\right\}$

Choosing $\diamond = 2$ we have ${c}_{2} \to 6 + 2 + {\square}_{1} = 9 \times {k}_{3}$
so ${\square}_{1} = 1$

Choosing $\diamond = 6$ we have ${c}_{2} \to 6 + 6 + {\square}_{2} = 9 \times {k}_{4}$
so ${\square}_{2} = 6$

The solutions are ${n}_{1} = 12132$ and ${n}_{2} = 12636$