Use the standard algorithm to find 3,135÷15.

How many times does 15 go into 31? – 2 times, with a remainder of 1. 2. How many times does 15 go into 13? – 0 times, with a remainder of 13. 3. Continue the division: – 15 into 135? – 9 times, with no remainder. Therefore, 3,135÷15=209.

The Correct Answer and Explanation is:

To solve the division problem (3,135 \div 15) using the standard algorithm, we proceed through several steps, breaking down the division to make it easier to understand.

Step 1: Setup the Division

We write (3,135) (the dividend) under the long division symbol and (15) (the divisor) outside of it.

Step 2: Divide the First Two Digits

We first look at how many times (15) can fit into the first part of the number, (31):

31 divided by 15 gives us 2 (since (15 \times 2 = 30)).
We write (2) above the division bar and subtract (30) from (31), which gives us a remainder of 1.

Step 3: Bring Down the Next Digit

Next, we bring down the next digit of the dividend, which is (3). Now, we are working with (13) (the remainder from the previous step plus the digit we just brought down):

15 into 13 goes 0 times (since (13 < 15)).
We write (0) above the division bar, and since (0) times (15) gives (0), we keep (13) as the new remainder.

Step 4: Bring Down the Next Digit Again

Now we bring down the next digit, which is (5), making it (135):

15 into 135 goes 9 times (since (15 \times 9 = 135)).
We write (9) above the division bar. Subtracting (135) from (135) gives us a remainder of 0.

Conclusion

Now we can summarize our findings:

The complete quotient is (209) with no remainder, so (3,135 \div 15 = 209).

Verification

To verify our answer, we can multiply (209) by (15):
[ 209 \times 15 = 3,135 ]
Since our product matches the original dividend, we confirm that the division is correct.

In summary, using the standard division algorithm, we systematically divided (3,135) by (15) and found that the answer is (209). Each step was crucial for breaking down the numbers, leading to the final result with clear verification through multiplication.