I'll Carry You Home

Clearly we have $\frac{1}{9} = 0.1111111111111111......$ $\frac{1}{99} = 0.010101010101010.........$ $\frac{1}{999} = 0.00100100100100100100100........$ 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..

$71$ is a prime number so without carry over the 71st digit is supposed to be $2$ .But $72 = 2^{3}*3^{2}$ 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 $72$ nd digit....

Hence the $71$ st digit of the number should be $2+1 = \boxed{3}$

First, let's list given fractions. $\dfrac{1}{9} = 0.11111111111 \cdots = 0.\overline{1}$ . $\dfrac{1}{99} = 0.01010101010101 \cdots = 0.\overline{01}$ . We find that this pattern continues on to infinity, with $\dfrac{1}{999} = 0.\overline{001}$ , $\dfrac{1}{9999} = 0.\overline{0001}$ , etc. Thus there will be $1$ added to the value of every place by $\dfrac{1}{9}$ , another $1$ added to the value of every other place by $\dfrac{1}{99}$ , etc. Since no number less than $100$ can have a factor greater than $100$ , and we are working with $71$ , and $71 < 100$ , we don't need to worry about the end fraction. We know that $71$ has $2$ factors, and $72$ has $12$ . The $1$ from the $12$ carries over, and the digit is $\fbox{3}$

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.

$0.1223242434262445262644283446282644492448282664303646284844322467482648\boxed{3}2246648305432444832....$

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