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Base-10 notation problem

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by cramya » Tue Oct 07, 2008 9:05 pm
Dint know any shortcut

Took 100 and subtracted 74 got 26
10^3 = 1000 - 74 = 926
10^4 = 10000-74 = 9926
10^5 = 100000 - 74 = 99926

Noticed a pattern where the last 2 digits are always going to be 26 and I got 2 9's lesser than avtual power 10 was raised to

eg: 10^5 I got 3 9's

So for 10^ 50 I will get 48 9's and the last 2 digits being 2 and 6

48*9+2+6
= 432 + 2 + 6
= 440

C)
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Re: Base-10 notation problem

by Morgoth » Tue Oct 07, 2008 11:36 pm
gmat009 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

OA is C

10^3 = 1000 [4 digits]
10^5 = 100000 [6 digits]

similarly,
10^50 = 1000....[51 digits]


1000 - 74 = 926
100000 - 74 = 99926

10^50 - 74 = 99999999......26

Therefore the sum of the digits = 9*48 = 432 + 2 + 6 = 440.


Hope this helps.
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by stop@800 » Wed Oct 08, 2008 1:55 am
cramya wrote:Dint know any shortcut
IMO there is no shortcut.
This is the best method you mentioned.

Took 100 and subtracted 74 got 26
10^3 = 1000 - 74 = 926
10^4 = 10000-74 = 9926
10^5 = 100000 - 74 = 99926

Noticed a pattern where the last 2 digits are always going to be 26 and I got 2 9's lesser than avtual power 10 was raised to

eg: 10^5 I got 3 9's

So for 10^ 50 I will get 48 9's and the last 2 digits being 2 and 6

48*9+2+6
= 432 + 2 + 6
= 440

C)
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by gmat009 » Wed Oct 08, 2008 7:32 am
Thanks everyone
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by niraj_a » Sat Dec 13, 2008 11:52 am
this is how i did it

74 = 10^2 - 26

so (10^50 - 10^2) - 26

10^2 (10^48) - 26

74 (10^48)

7+4 = 11. 440 is divisible by 11.
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