The smallest 5 digit number exactly divisible by 41 is:

Answers (1)

We know that smallest 2 digit number

is 10.

Smallest three digit number is 100.

Similarly smallest five digit number is

10000.

Here we use,
Division algorithm:

Dividend = divisor × quotient + remainder

Dividend = 10000

Divisor = 41

10000 = 41 × 243 + 37

To find required number we have

add number = divisor - remainder

= 41 - 37

= 4 to 10000

Therefore, Smallest 5 digit number exactly divisible by 41 is 10004.

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