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10^50 - 74

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10^50 - 74 [#permalink] New post 18 May 2010, 22:08
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If 10^50 - 74 is written as an integer in a base 10 notation.What is the sum of the digits in that integer?

a. 424
b. 433
c. 440
d. 449
e. 467
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Re: 10^50 - 74 [#permalink] New post 18 May 2010, 22:38
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C. 440

10^x - 74 --> last 2 digits are always 2, 6.

10^2 - 74 = 26

10^3 - 74 = 926

10^4 - 74 = 9926 and so on....

If x > 1,
the sum of the digits --> (x-2) * 9 + 2 + 6. hence, (50-2) * 9 + 8 --> 440.
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Re: 10^50 - 74 [#permalink] New post 18 May 2010, 23:41
nsp007 wrote:
C. 440

10^x - 74 --> last 2 digits are always 2, 6.

10^2 - 74 = 26

10^3 - 74 = 926

10^4 - 74 = 9926 and so on....

If x > 1,
the sum of the digits --> (x-2) * 9 + 2 + 6. hence, (50-2) * 9 + 8 --> 440.


great approach but why are you using x here ? as you used ..."If x > 1," ..
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Re: 10^50 - 74 [#permalink] New post 19 May 2010, 00:06
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dimitri92 wrote:
If 10^50 - 74 is written as an integer in a base 10 notation.What is the sum of the digits in that integer?

a. 424
b. 433
c. 440
d. 449
e. 467


C. 440
another approach is:
We know that 10^50 is ending 00, so 10^50-74=9....9926
total number of digits in 10^50-74 is 50, or 48 digits of 9 and two digits 2 and 6.
answer choice is 48*9+8=440

plugging numbers:
let represent the sum of the digits in that integer as Y, with the reminder 8, we can represent it in form Y=X*9+8, where X number of digits in 10^50-74 and 8=2+6.

Start with C and than move to B or D.

B. 433=X*9+1, X=48
C. 440=X*9+8, X=48 - correct as we have the reminder 8 and 48 number of digits (50-2), 2 digits are 26.
D. 449=X*9+8, X=49



Personally, I like NSP007's approach. My approaches are easy to comprehend.
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Re: 10^50 - 74 [#permalink] New post 19 May 2010, 10:00
great approach but why are you using x here ? as you used ..."If x > 1," ..


I meant to say that only when x > 1, for 10^x - 74.. the last 2 digits are 2, 6.
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Re: 10^50 - 74 [#permalink] New post 09 Jun 2010, 00:03
Hey Pkit,
Please explain your approach. How do you know "total number of digits in 10^50-74 is 50, or 48 digits of 9 and two digits 2 and 6"?
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Re: 10^50 - 74 [#permalink] New post 09 Jun 2010, 05:38
bibha wrote:
Hey Pkit,
Please explain your approach. How do you know "total number of digits in 10^50-74 is 50, or 48 digits of 9 and two digits 2 and 6"?

Approach is easy
Just look at the following example, what is the sum of digits of 10^3-5 ?
you know that 10^3=1000 (thus 10^50 is a figure that begins with 1 anf has 50 zeros) and 1000-5=995, so I have two "9" and one "5". the sum is 9+9+5=23

kudos! :wink:
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Re: 10^50 - 74 [#permalink] New post 09 Jun 2010, 09:10
Great explanation. Thanks
Re: 10^50 - 74   [#permalink] 09 Jun 2010, 09:10
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