If 10^50 - 74 is written as an integer in a base 10 notation.What is the sum of the digits in that integer?
a. 424
b. 433
c. 440
d. 449
e. 467
a. 424
b. 433
c. 440
d. 449
e. 467
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10^50-74
Source: Beat The GMAT — Data Sufficiency |
The answer is C.
10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...
so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...
so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
Can you explain the step marked with Bold.prindaroy wrote:The answer is C.
10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...
so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
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The last two digits are 2 and 6.gmat740 wrote:Can you explain the step marked with Bold.prindaroy wrote:The answer is C.
10 ^ 50 = 51 digits, i.e 10^2 = 100 = 3 digits, 10^3 = 1000 = 4 digits, etc...
so 10.........-74 = 999999........26 with 50 digits in total. So now, 2+6+(9*48)
= 440
The first 48 digits are 9.
So when we sum all the digits, we get 48 9s, one 2 and one 6 to gives us 48*9 + 2 + 6
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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