The smallest 5-digit number is 10000. To find the smallest 5-digit number that is exactly divisible by 41, we need to divide 10000 by 41 and round up to the nearest whole number.
First, perform the division:
10000 ÷ 41 ≈ 243.9024
Now, we round this up to the next whole number, which is 244.
Next, we multiply 244 by 41 to get the smallest 5-digit number that is divisible by 41:
244 × 41 = 10004
Therefore, the smallest 5-digit number that is exactly divisible by 41 is 10004.