Question

The smallest 5 digit number exactly divisible by 41 is:

Answer 1

Hi,

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,

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Division algorithm:

Dividend = divisor × quotient +

remainder

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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.

Verification:
__________


10004 = 41 × 244 + 0

I hope this will useful to you.

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Answer 2

Answer:

10004 is the answer of this question