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
Base-10 notation problem
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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)
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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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.
IMO there is no shortcut.cramya wrote:Dint know any 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)