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

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

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New post 22 Sep 2015, 02:50
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70% (01:52) correct 30% (02:12) wrong based on 207 sessions

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Re: What is the sum of the digits of integer k, if k = (10^40- 46)  [#permalink]

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New post 22 Sep 2015, 04:41
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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
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Re: What is the sum of the digits of integer k, if k = (10^40- 46)  [#permalink]

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New post 22 Sep 2015, 11:51
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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
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Re: What is the sum of the digits of integer k, if k = (10^40- 46)  [#permalink]

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New post 27 Sep 2015, 11:11
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.


VERITAS PREP OFFICIAL SOLUTION:

While this may look like a monster problem, it’s really just one of arithmetic. 10^40 is an insanely large number, but conceptually it’s not much different from 10^3 (i.e. 1000). If you test this relationship with a few small numbers, you can get a good look at what k will look like. For example:

\(10^2 -46 = 100-46 = 54\)
\(10^3 -46 = 1000-46 = 954\)
\(10^4 -46 = 10000-46 = 9954\)

Do you see the pattern? Every time we add one to the exponent, we add another 9 to the solution. And the number of digits in the solution is always the same as the exponent itself. So for this problem, where the exponent is 40, k will have 40 digits: a 5, a 4, and the other 38 are 9s. And since 5 + 4 is 9, then really we’re just adding up 39 9s. And 39*9 is 351 (or you can just see that it will end in a 1, and only A matches).
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Re: What is the sum of the digits of integer k, if k = (10^40- 46)  [#permalink]

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New post 22 Sep 2018, 05:14
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