Divisibility ?

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Divisibility ?

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Answer 1 · 2016-06-11T19:45:18

$n_{1} = 12132$ $n_1 = 12132$ and $n_{2} = 12636$ $n_2 = 12636$

Explanation:

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

We have

$c_{1} \to 3 \times 10 + ⋄ = 4 \times k_{1}$ $c_1->3 xx 10 + diamond = 4 xx k_1$
$c_2->1+2+square+3+diamond = 9 xx k_2$

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

From $c_1-> diamond in {2,6}$

Choosing $diamond = 2$ we have $c_2->6+2+square_1=9 xx k_3$
so $square_1=1$

Choosing $diamond = 6$ we have $c_2->6+6+square_2=9 xx k_4$
so $square_2 = 6$

The solutions are $n_1 = 12132$ and $n_2 = 12636$

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Divisibility ?

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