What is the sum of the digits of integer k, if k = (10^50- 50) Select:
A) 337  B) 437  C) 457  D) 537  E) 412
There are 51 digits in 10^50   When we subtract 50 from it, there will be 50 digits left.   10^50 can be written as 9999999....(50 times) + 1   So,   10^50 - 50 = 9999999....(50 times) + 1 - 50 = 9999999....(50 times) - 49   Consider the last 2 digits,   99-49= 
<gadget id="calculator">99-49</gadget>
<output>50</output>
50   The last 2 digits will be 50.   And our number would be 99999......99950 with 2 less 9s after subtraction.   Number of 9s left are 48 and the last two digits are 50   The sum of the digits will be     (48*9) + 5 + 0= 
<gadget id="calculator">  (48*9) + 5 + 0</gadget>
<output>437</output>
437   Answer:- 437
<result>B</result>
What is the sum of the digits of integer k, if k = (10^50- 50)