Base System

Given the 4 digit number :

Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.

Let the number be abcd.

Given that a+b+c = 14. (1)

                 b+c+d = 15. (2)

                 c = d+4. (3).

In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands place in order to maximize the value of the number. b, c, and d are less than 9 each as they are single-digit numbers.

Substituting (3) in (2) we have b+d+4+d = 15, b+2*d = 11.  (4)

Subtracting (2) and (1) : (2) - (1) = d = a+1.   (5)

Since c cannot be greater than 9 considering c to be the maximum value 9 the value of d is 5.

If d = 5, using d = a+1, a = 4.

Hence the maximum value of a = 4 when c = 9, d = 5.

Substituting b+2*d = 11. b = 1.

The highest four-digit number satisfying the condition is 4195

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.

$$(10^{50} + 10^{25} - 123)$$ can be written as $$10^{50} + (10^{25}-123)$$

$$1000-123$$ gives $$877$$. 
$$10000-123$$ gives $$9877$$.
$$100000-123$$ gives $$99877$$.
$$1000000-123$$ gives $$999877$$.
And so on.

Considering $$10^n-123$$, which has a total of $$n$$ digits; in general, of the $$n$$ total digits; the digit $$7$$ is repeating $$2$$ times, digit $$8$$ is repeating $$1$$ time, and digit $$9$$ is repeating $$(n-3)$$ times.

Also, when added to $$10^{50}$$, the cumulative increase in the sum of the digits of the resultant number would be $$1$$, which would come from the leftmost digit of the result. All other newly added digits will be zero.

Thus, the sum of the digits of the result will be given by;

$$[7*2] + [1*8] + [(25-3)*9] + [25*0] + [1*1] = 14+8+198+1 = 221$$

The correct answer is option B.