Test the divisibility of the following numbers by $7$ .
$94563$

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Solution

A number is divisible by $7$ if the difference between twice the last digit and number formed by other digits is either $0$ or a multiple of $7$
In the given number $94563$ last digit is $3$ . Twice of $3$ is $6$ and the number formed by other digit is $9456$ .
Diffierence $=$ $9456 - 6 = 9450$ , which is divisible by $7$ .
Hence, the number $94563$ is divisible by $7$ .

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