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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]  30 Jun 2005, 18:32
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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
Joined: 30 May 2005
Posts: 374
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[#permalink]  30 Jun 2005, 18:44
10^50 is 1 followed by 50 zeros.

10^50 - 74 = 999...926

The 9s occur 50-2 = 48 times.

Sum of digits = 48*9+6+2 = 440

C
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Re: PS [#permalink]  30 Jun 2005, 18:48
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

I got C.

10^50 has 1 followed by 50 zeroes.
10^50 - 74 has 48 9's with 26 at the last two places.
sum of the digits = 48*9 + 8 = 440.

HMTG.
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[#permalink]  30 Jun 2005, 18:50
OA is C.

Thanks guys. I was trying 48*9 + 26 and coming up with no match, I guess I must call it a day.
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Re: PS [#permalink]  30 Jun 2005, 18:51
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. Explanations please

10^2-74=100-74 = 26
10^3-74=1,000-74 = 926
10^4-74=10,000-74 = 9926
.
.
.
10^50-74= fortyeight 9's and 26= 48X9+26= the last digit is zero so it should be 440.
Re: PS   [#permalink] 30 Jun 2005, 18:51
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# If 10^50 - 74 is written as an integer in base 10 notation,

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