Question:
Given that the number $\overline{35 \alpha 64}$ is divisible by 3 , where $\alpha$ is a digit, what are the possible values of $\alpha$ ?
Solution:
It is given that $\overline{35 a 64}$ is a multiple of 3 .
$\therefore(3+5+a+6+4)$ is a multiple of 3 .
$\therefore(a+18)$ is a multiple of 3 .
$\therefore(a+18)=0,3,6,9,12,15,18,21 \ldots$
But $a$ is a digit of number $\overline{35 a 64}$. So, $a$ can take value $0,1,2,3,4 \ldots 9$.
$a+18=18 \Rightarrow \mathrm{a}=0$
$a+18=21 \Rightarrow \mathrm{a}=3$
$a+18=24 \Rightarrow \mathrm{a}=6$
$a+18=27 \Rightarrow \mathrm{a}=9$
$\therefore a=0,3,6,9$
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