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N = 1111 1 1998 times N=\underbrace{1111\ldots1}_{\text{1998 times}}

Find the 100 0 th 1000^\text{th} digit after decimal point of N \sqrt{N} .


The answer is 1.

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2 solutions

Lam Nguyen
Sep 5, 2016

111111111 111111111 consists of 9 ones. 9 / 9 = 1 9/9 =1 Odd Value

111111111 = 12345679 9 111 111 111= 12345679 *9

12345679 9 = 10540.92553 ) \sqrt{12345679} * \sqrt{9} = 10540.92553)

111111111111111111 111111111111111111 consists of 18 ones 18 / 9 = 2 18/9=2 Even Value

111111111111111111 = 12345679012345679 9 111 111 111 111 111 111= 12345679012345679 *9

12345679012345679 9 = 111111111.1 3 = 333333333.3 ) \sqrt{12345679012345679} * \sqrt{9} = 111111111.1 * 3 = 333333333.3)

Therefore for odd value of 9, it is not possible to predict the digit. However, for every even value of 9, all the digits are three.

1998 / 9 = 222 1998/9=222 Even Value Therefore all the digits will be 3

I think the answer should be 3 and not 1

Abhishek Alva
Sep 1, 2016

please who ever has solved the problem sincerely kindly give the solution

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