Consider the following series written in form of a decimal A = 9 1 + 9 9 1 + 9 9 9 1 + . . . . + 1 0 1 0 0 − 1 1
Find the digit in the 7 1 st place after the decimal point.
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sir, even though i got it, but your explanation is phew!!!!!!!!!!!
But we start adding from the right hand side,what if there are more carry overs affecting 12 which we got in the 72nd column?
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That's a good point but here you can intuitively see by inspection that there is no such case....
I forget about that -_-
Exactly Same Way.
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Same way....
1234567890+×÷=%/*€£@$!#:;&_()-'",.?😰
No, 9 9 1 = 0 . 0 1 , not 0 . 1 0 . Similar with 9 9 9 1 . But the answer is still luckily correct.
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I just made a typographical error which has Nothing to do with the concept....... or the answer for that matter
same as mine. Hahahaha
First, let's list given fractions. 9 1 = 0 . 1 1 1 1 1 1 1 1 1 1 1 ⋯ = 0 . 1 . 9 9 1 = 0 . 0 1 0 1 0 1 0 1 0 1 0 1 0 1 ⋯ = 0 . 0 1 . We find that this pattern continues on to infinity, with 9 9 9 1 = 0 . 0 0 1 , 9 9 9 9 1 = 0 . 0 0 0 1 , etc. Thus there will be 1 added to the value of every place by 9 1 , another 1 added to the value of every other place by 9 9 1 , etc. Since no number less than 1 0 0 can have a factor greater than 1 0 0 , and we are working with 7 1 , and 7 1 < 1 0 0 , we don't need to worry about the end fraction. We know that 7 1 has 2 factors, and 7 2 has 1 2 . The 1 from the 1 2 carries over, and the digit is 3
involving the carrys, the 72nd row will have a sum of 12, while the 71st row will have 2, and therefore the digit that will appear at the 71st place will be 2 + 1(carry) = 3
nice problem got tricked the first time, overlooked 72 then, later corrected it! Enjoyed solving it!
Using the result from a previous quiz problem, counting the decimals ... QED.
0 . 1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6 4 4 2 8 3 4 4 6 2 8 2 6 4 4 4 9 2 4 4 8 2 8 2 6 6 4 3 0 3 6 4 6 2 8 4 8 4 4 3 2 2 4 6 7 4 8 2 6 4 8 3 2 2 4 6 6 4 8 3 0 5 4 3 2 4 4 4 8 3 2 . . . .
1/9 is 0.11111111111 1/99 is 0.10101010 1/999 is0.001001001 And so on..... If u notice there is pattern her in recurring of zeros...so 71 th letter has to be 1+1+1.....(understand the pattern)....so 3
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Clearly we have 9 1 = 0 . 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 . . . . . . 9 9 1 = 0 . 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 . . . . . . . . . 9 9 9 1 = 0 . 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 . . . . . . . . So clearly the digit in the k place will be equal to the number of factors of that number.
But we also have to consider the carry overs..
7 1 is a prime number so without carry over the 71st digit is supposed to be 2 .But 7 2 = 2 3 ∗ 3 2 has 12 factors!!!And hence will get carried over and added to the 71st digit!!By inspection we can say that there is no significant carry over after the 7 2 nd digit....
Hence the 7 1 st digit of the number should be 2 + 1 = 3