Bunuel wrote:

What is the sum of the digits of integer k, if k = (10^40- 46)

(A) 351

(B) 360

(C) 363

(D) 369

(E) 378

Kudos for a correct solution.

There are 41 digits in 10^4

When we subtract 46 from it, there will be 40 digits left.

10^4 can be written as 9999999....(40 times) + 1

So,

10^40 - 46 = 9999999....(40 times) + 1 - 46 = 9999999....(40 times) -45

Consider the last 2 digits,

99-45 = 54

The last 2 digits will be 54.

And our number would be 99999......99954 with 2 less 9s after subtraction.

Nuber of 9s left are 38 and the last two digits are 54

The sum of the digits will be

(38*9) + 5 + 4 =351

Answer:- A