What is the sum of the digits of integer k, if k = (10^45- 46) Select.
A) 356	B) 366	C) 363	D) 396	E) 408
There are 46 digits in 10^45   When we subtract 46 from it, there will be 45 digits left.   10^45 can be written as 9999999....(45 times) + 1   So,   10^45 - 46 = 9999999....(45 times) + 1 - 46 = 9999999....(45 times) -45   Consider the last 2 digits,   99-45= 
<gadget id="calculator">99-45</gadget>
<output>54</output>
54   The last 2 digits will be 54.   And our number would be 99999......99954 with 2 less 9s after subtraction.   Number of 9s left are 43 and the last two digits are 54   The sum of the digits will be     (43*9) + 5 + 4= 
<gadget id="calculator">  (43*9) + 5 + 4</gadget>
<output>396</output>
396   Answer:- 396
<result>D</result>
What is the sum of the digits of integer k, if k = (10^45- 46)