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Mathematics 26 Online
OpenStudy (anonymous):

The number of digits in the sum 100+ 100^2 + 100^3 + ….+ 100^2011 is a) 4023 b) 4022 c) 4024 d) None of these

OpenStudy (anonymous):

B

OpenStudy (anonymous):

How bhaiya?

OpenStudy (anonymous):

ah 100 + 100^2 = 100 + 10000 = 10100 100 + 10000 + 1000000 = 10100100

OpenStudy (anonymous):

so 2 zero position and that makes 2X2011 =2022 and plus 1 position the answer is 4023

OpenStudy (anonymous):

ah I meant option A not B read the options wrong

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