Find the smallest 5 digit number exactly divisible by 41.
We use the division algorithm to find the smallest 5 digit number exactly divisible by 41.
Answer: The smallest 5-digit number exactly divisible by 41 is 10004.
Let's use the condition given in the question to find the required number.
We already know the smallest 5-digit number is 10000.
We know that according to the division algorithm:
Dividend = Divisor × Quotient + Reminder
Now, once we divide 10000 by 41, we get:
10000 = 41 × 243 + 37
Since 37 is the remainder when 10000 is divided by 41 it means that if we add 4 more to 37, we get a number perfectly divisible by 41.
Adding 4 to 37 means adding 4 to 10000, that is, the number 10004 will be completely divisible by 41.
Therefore, the smallest 5-digit number exactly divisible by 41 is 10004.