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If 2 1 1 equals the repeating decimal 0 . 0 4 7 6 1 9 0 4 7 6 1 9 0 . . . , what is the 5 1 s t digit after the decimal point of the repeating decimal?
The answer is 7.
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4 solutions
I don't understand this
Instead of 0 . 0 4 7 6 1 9 0 4 7 6 1 9 0 . . . let`s look at 0 . 4 7 6 1 9 0 4 7 6 1 9 0 . . . It is clear from looking at the second repeating decimal that the 6 n t h decimal place will be 0 which will be at the at the ( 6 n t h + 1 ) t h place in the original repeating decimal .The 4 8 t h d.p of the 2 n d decimal will be 0 so the 4 9 t h d.p of the first decimal will be 0 . 5 0 t h d.p of the first decimal will be 4 . 5 1 s t d.p of the first decimal will be 7 .
Terrible problem. If 1/21 = (a number with 13 digits after the decimal) then the 51st digit will be the second to last number (4 x 13 = 52, 52 -1) in the original sequence (9). That answer is incorrect because the question is written in such a way as to provide confusion instead of clarity.
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There are six digits that repeat. The last 0 was added to indicate that the sequence keeps restarting. Fairly obvious if you just look at the numbers.
The way this was Wooten is purely worded the last 0 needs to be omitted. ... it makes it sounds like there it's supposed to be a double 0.... it does not repeat what it says. ... delete the last 0
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Agreed, as written, there are 13 digits after the decimal that repeat.
N otice that the digits are 0 , 4 , 7 , 6 , 1 , 9 are repeated again and again . they are total 6 in number so ==> 5 1 = 3 ( m o d 6 ) the third number in the above set is 7 so 7 is our answer .
51 is not equal to 3(mod6) it is congruent to 3(mod6). There is a massive difference!
But it is the same way as I did it.
Instead of 0.0476190476190... let`s look at 0.476190476190... It is clear from looking at the second repeating decimal that the 6nth decimal place will be 0 which will be at the (6nth+1)th place in the original repeating decimal.The 48th d.p of the 2nd decimal will be 0 so the 49th d.p of the first decimal will be 0.50th d.p of the first decimal will be 4. 51st d.p of the first decimal will be 7.
Since there are 6 repeating digits, we can divide 51 by modulo 6, giving 8 (nearest muliple of 6 being 6 by 8 = 48) with a remainder of 3. The third digit of the repeating series is 7.