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

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Joined: 06 Nov 2008
Posts: 53

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

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New post 11 Dec 2008, 15:09
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

Kudos [?]: 6 [0], given: 0

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Posts: 1863

Kudos [?]: 627 [0], given: 32

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Re: integer [#permalink]

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New post 11 Dec 2008, 15:11
C. 440.

Whenever there are large exponents like this, I like to use a smaller exponent and extract a rule from that simply because it's easier to use smaller numbers.

if we use 1000 (which is 10^3) we have 1000 - 74 = 926. So we know we'll have 2+6 for 8, but how many 9's will we have with the exponent being 50? well, if used 100 (10^2), we get 100 - 74 = 26, so here there are zero 9's. So let x represent the exponent value, we know that x - 2 will be the number of 9's. so 9(x-2) + 2 + 6 where x = the value of the exponent.

If x = 50, then 50 -2 = 48...48*9 = 432 + 2 + 6 = 440

haichao 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

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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Kudos [?]: 627 [0], given: 32

Re: integer   [#permalink] 11 Dec 2008, 15:11
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If 10^50 74 is written as an integer in base 10 notation,

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