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08 Mar 2022, 21:44
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Let x = 0.12345678910111213. . . 998999, where the digits are obtained by writing the integers 1 through 999 in that order. The 1983rd digit to the right of the decimal point is?
A. 2
B. 3
C. 5
D. 7
E. 8
A. 2
B. 3
C. 5
D. 7
E. 8
09 Mar 2022, 00:29
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No. of single digits (1 to 9) are : 1 x 9 = 9
No. of double digits (10 to 99) are : 2 x 90 = 180
No. of triple digits (100 to 999) are : 3 x 900 = 2700
So, we will take total triple digit number as 3 x n = 3n ,because the 1983rd number will lie somewhere between this range.
1x9+2x90+3xn = 1983
3n=1983-9-180 = 1794
n=598
and we know that triple digit number starts from 100, so the 1983rd term will be 597, so the answer is 7.
(100,101,102,103.................593,594,595,596,597,598.........998,999
No. of double digits (10 to 99) are : 2 x 90 = 180
No. of triple digits (100 to 999) are : 3 x 900 = 2700
So, we will take total triple digit number as 3 x n = 3n ,because the 1983rd number will lie somewhere between this range.
1x9+2x90+3xn = 1983
3n=1983-9-180 = 1794
n=598
and we know that triple digit number starts from 100, so the 1983rd term will be 597, so the answer is 7.
(100,101,102,103.................593,594,595,596,597,598.........998,999
09 Mar 2022, 06:43
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so basically the question involves counting of digits. in this case
1- 9= 9 digits
10-99= 90* 2 = 180 digits
100- 999= 900*3 = 2700 digits (1983rd number will lie in this range)
so 1983-9-180= 1794
dividing 1794 by 3 we get 598. Ans will not be 8, it will be 7 since the range begins inclusive of 100.
1- 9= 9 digits
10-99= 90* 2 = 180 digits
100- 999= 900*3 = 2700 digits (1983rd number will lie in this range)
so 1983-9-180= 1794
dividing 1794 by 3 we get 598. Ans will not be 8, it will be 7 since the range begins inclusive of 100.
