Question

Easy method to find the smallest 5 digit number divisible by 41

Answer 1

➡HERE IS YOUR ANSWER⬇

The smallest 5-digit number is = 10000.

We see that there are zeroes at the end of 10000, which makes the number the smallest.

To find the smallest 5-digit number, divisible by 41, we need to put zeroes at the end.

If we multiply 41 by 1000, we get a number

= 41 × 1000

= 41000, which is divisible by 41.

Hence, the smallest required number is 41000.

⬆HOPE THIS HELPS YOU⬅