The smallest 5-digit number divisible by 41 is ___

Options

A. 10054
B. 10041
C. 1004
D. 4100

Answer

To find the smallest 5-digit number that is divisible by 41, we can follow these steps:

### Step 1: Identify the smallest 5-digit number
The smallest 5-digit number is 10,000.

### Step 2: Divide 10,000 by 41
We need to determine how many times 41 fits into 10,000. We perform the division:
\[ 10,000 \div 41 \approx 243.9024 \]
This means 41 fits into 10,000 approximately 243 times.

### Step 3: Calculate the product of 41 and 243
Now, we multiply 41 by 243 to find the largest multiple of 41 that is less than 10,000:
\[ 41 \times 243 = 9,963 \]

### Step 4: Calculate the remainder
Next, we find the remainder when 10,000 is divided by 41:
\[ 10,000 - 9,963 = 37 \]
This means that 10,000 is 37 more than the nearest lower multiple of 41.

### Step 5: Find the next multiple of 41
Since we need the smallest 5-digit number that is divisible by 41, we need to add the difference to 10,000 to reach the next multiple of 41:
\[ 10,000 + (41 - 37) = 10,000 + 4 = 10,004 \]

### Conclusion
Thus, the smallest 5-digit number divisible by 41 is:
\[ \text{Answer: } 10,004 \]

---