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If 10^50 74 is written as an integer in base 10 notation,

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If 10^50 74 is written as an integer in base 10 notation, [#permalink]

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New post 01 Aug 2006, 11:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

A. 424
B. 433
C. 440
D. 449
E. 467
Senior Manager
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Re: base 10 notation - problem [#permalink]

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New post 01 Aug 2006, 12:05
positive_energy wrote:
If 10^50 – 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?

A. 424
B. 433
C. 440
D. 449
E. 467

C it is.

10^50 = 10........0 (the number of 0 is 50)
10^50 - 74 = 9.......926 (the number of 9 is 48)

9*48+2+6 = 440
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New post 01 Aug 2006, 12:14
C

10^50 have 50 zeros
10^50 - 74 has 48 9s and 26 in the end i.e 9999......48 times......26
Sum = 48 *9 + 2+6 = 440
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New post 02 Aug 2006, 03:33
C 440

1 ............... 49th 50th
1 0 0
- 7 4
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9999.... 2 6

There will be 48 .....9's
hence sum = 48*9 + 2+6 = 432 + 8 = 440
  [#permalink] 02 Aug 2006, 03:33
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If 10^50 74 is written as an integer in base 10 notation,

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