Answer :

We want to compute
[tex]$$
6510 \div 31.
$$[/tex]

Step 1: Divide the first part.

Start with the first two digits of 6510, which is 65, since 31 is greater than the first digit 6. Determine how many times 31 fits into 65:
[tex]$$
31 \times 2 = 62.
$$[/tex]
Since 62 is less than 65 and 31 cannot fit 3 times (because [tex]$31 \times 3 = 93$[/tex] is too large), we use 2. Write 2 as the first digit of the quotient. Now subtract:
[tex]$$
65-62=3.
$$[/tex]

Step 2: Bring down the next digit.

Bring down the next digit (1) next to the remainder 3. This gives us:
[tex]$$
31 \quad \text{(from the 3 and 1)}.
$$[/tex]
Now, determine how many times 31 fits into 31:
[tex]$$
31 \times 1 = 31.
$$[/tex]
Write 1 as the next digit of the quotient. Subtract:
[tex]$$
31 - 31 = 0.
$$[/tex]

Step 3: Bring down the final digit.

The final digit, 0, is brought down next to 0, giving 0. Since 31 cannot fit into 0, write a 0 as the last digit of the quotient:
[tex]$$
31 \times 0 = 0.
$$[/tex]
Subtracting, we still have [tex]$0$[/tex] as the remainder.

Conclusion:

The quotient obtained from the long division is
[tex]$$
210,
$$[/tex]
and the remainder is 0. This means that
[tex]$$
6510 = 31 \times 210.
$$[/tex]

Thus, the final answer is: [tex]$$\boxed{210}.$$[/tex]

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