What is the sum of the digits of integer k, if k = (10^40- 79)  Choose the correct choice from the following answers:
A) 345 B) 340 C) 343 D) 349 E) 338
There are 41 digits in 10^40   When we subtract 79 from it, there will be 40 digits left.   10^40 can be written as 9999999....(40 times) + 1   So,   10^40 - 79 = 9999999....(40 times) + 1 - 79 = 9999999....(40 times) -78   Consider the last 2 digits,   99-78= 
<gadget id="calculator">99-78</gadget>
<output>21</output>
21   The last 2 digits will be 21.   And our number would be 99999......99921 with 2 less 9s after subtraction.   Number of 9s left are 38 and the last two digits are 21   The sum of the digits will be     (38*9) + 2 + 1= 
<gadget id="calculator">  (38*9) + 2 + 1</gadget>
<output>345</output>
345   345nswer:- 345
<result>A</result>
What is the sum of the digits of integer k, if k = (10^40- 79)