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# If 10^(50) - 74 is written as an integer in base 10

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Senior Manager
Joined: 08 Aug 2005
Posts: 251
If 10^(50) - 74 is written as an integer in base 10 [#permalink]

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05 May 2006, 23:18
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
VP
Joined: 29 Dec 2005
Posts: 1341
Re: sum of the digits [#permalink]

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16 May 2006, 13:40
getzgetzu 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

10^2-74 = 26
10^3-74 = 926 = one (3-2) 9's
10^4-74 = 9926 = two (4-2) 9's
.
.
.
.
.
.
10^50-74 = 9999....9926 = forty-eight (50-2) 9's

sum = 48 x 9 + 26 = 458
Manager
Joined: 07 Sep 2004
Posts: 60

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17 May 2006, 05:14
Agree with the reasoning

there are 48 of a nine
there is a 2
there is a 6

so sum = 48*9 + 2+ 6 = 440
CEO
Joined: 20 Nov 2005
Posts: 2894
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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17 May 2006, 07:09
10^50 -74 = 9999..........26. There are 48 9's and 2 and 6.

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

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

VP
Joined: 29 Dec 2005
Posts: 1341
Re: sum of the digits [#permalink]

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17 May 2006, 11:25
getzgetzu 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

10^2-74 = 26
10^3-74 = 926 = one (3-2) 9's
10^4-74 = 9926 = two (4-2) 9's
.
.
.
.
.
.
10^50-74 = 9999....9926 = forty-eight (50-2) 9's
sum = 48 x 9 + 2+6 = 440

ufffffffffffffffffff
Director
Joined: 10 Oct 2005
Posts: 718

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17 May 2006, 12:01
ps_dahiya wrote:
10^50 -74 = 9999..........26. There are 48 9's and 2 and 6.

Sum of digits = 9*48 +2+6 = 440
same approach but silly mistake in calculations
_________________

IE IMBA 2010

17 May 2006, 12:01
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