For a 4-digit number (greater than 1000), sum of the digits in the thousands, hundreds, and tens places is 15. Sum of the digits in the hundreds, tens, and units places is 16. Also, the digit in the tens place is 6 more than the digit in the units place. The difference between the largest and smallest possible value of the number is

Let the thousands, hundreds, tens, and units digits of the number be $$a$$, $$b$$, $$c$$, and $$d$$, respectively.

The sum of the digits in the thousands, hundreds, and tens places is 15;  $$a+b+c = 15$$ .....(1)

The sum of the digits in the hundreds, tens, and units places is 16;  $$b+c+d = 16$$ .....(2)

Solving equations (1) and (2), we get $$d = a+1$$.

The digit in the tens place is 6 more than the digit in the units place;  $$c=d+6 = a+1+6 = a+7$$ .....(3)

Substituting the value of $$c$$ in terms of $$a$$ in equation (1), we get;  $$a+b+a+7 = 15$$  or  $$b=8-2a$$

Thus, the number is:  $$\underline{a}\text{ }\underline{8-2a}\text{ }\underline{a+7}\text{ }\underline{a+1}$$

We know that since the number is a four digit number, $$a\geq 1$$. But all the four digits have to be less than or equal to $$9$$ but greater than or equal to $$0$$, thus,

$$8-2a\geq 0$$  gives  $$a\leq 4$$  and  $$a+7\leq 9$$  gives  $$a\leq 2$$.

Thus, $$a$$ can be $$1$$ or $$2$$.

The two possible values for the number, therefore, are, $$1682$$ and $$2493$$, when we put $$a=1$$ and $$a=2$$, respectively.

The difference between the only two possible values is, $$2493 - 1682 = 811$$

Option A is the correct answer.