Question

The smallest $5$ digit number exactly divisible by $41$ is:

Answer 1

The smallest five-digit number is $10000.$

We will use Division algorithm

We know that

Dividend = divisor × quotient + remainder

Dividend = $10000.$

Divisor = $41$

$10000=41\times 243+37$

To find required number we have to

Add numbers$=$Divisor $-$ remainder

$=41–37$
$=4$to $10000$

Therefore, Smallest 5 digit number exactly divisible by $41\mathrm{is}10004.$