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22 Sep 2015, 02:50
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:50) correct
24%
(02:06)
wrong
based on 389
sessions
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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.
(A) 351
(B) 360
(C) 363
(D) 369
(E) 378
Kudos for a correct solution.
22 Sep 2015, 04:41
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Bunuel
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
22 Sep 2015, 11:51
Kudos
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(10^40- 46) is nothing but 999... (38 of them) followed by 54
so that adds to (9*38)+5+4 = 351
Answer is A
so that adds to (9*38)+5+4 = 351
Answer is A
