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Question

The smallest 5 digit number exactly divisible by 41 is:


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Solution

The smallest five-digit number is 10000.

We will use Division algorithm

We know that

Dividend = divisor × quotient + remainder

Dividend = 10000.

Divisor = 41

10000=41×243+37

To find required number we have to

Add numbers=Divisor - remainder

=4137
=4to 10000

Therefore, Smallest 5 digit number exactly divisible by 41is10004.


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