Call the decimal \(x\), so \(x = 0.\dot{ 5 }\)

One digit repeats, so multiply by 10. That moves the repeating block exactly one block to the left:

\(10x = 5.\dot{ 5 }\)

Both numbers now have identical digits after the point, so subtracting removes the recurring tail completely:

\(10x - x = 5.\dot{ 5 } - 0.\dot{ 5 }\), which gives \(9x = 5\)

\(x = \frac{ 5 }{9}\)

That fraction is already in its lowest terms.

\(0.\dot{ 5 } = \frac{ 5 }{ 9 }\)

The dots mark the first and last digits of the repeating block, so the block here is 65 and the decimal is \(0.656565\ldots\)

Call it \(x\). Two digits repeat, so multiply by 100 this time:

\(100x = 65.\dot{ 6 }\dot{ 5 }\)

Subtracting removes the tail:

\(100x - x = 65\), which gives \(99x = 65\)

\(x = \frac{ 65 }{99}\)

That fraction is already in its lowest terms.

\(0.\dot{ 6 }\dot{ 5 } = \frac{ 65 }{ 99 }\)

Divide 13 by 15.

\(13 \times 10 = 130\), and \(130 \div 15 = 8\) remainder \(10\)

That gives the first decimal digit: \(0.8\)

Bring down a zero and divide again: \(10 \times 10 = 100\), and \(100 \div 15 = 6\) remainder \(10\)

The remainder has come back to \(10\), the same one as before, so from here the division just repeats and the digit \(6\) recurs forever.

\(\frac{ 13 }{ 15 } = 0.8\dot{ 6 }\)

Only the 5 repeats. The 8 sits between the point and the start of the repeating block, so multiplying by 10 alone will not line the tails up.

Call the decimal \(x\), so \(x = 0.8\dot{ 5 }\)

\(10x = 8.\dot{ 5 }\)

\(100x = 85.\dot{ 5 }\)

Those two have the same recurring tail, so subtract them rather than subtracting \(x\):

\(100x - 10x = 85 - 8\), which gives \(90x = 77\)

\(x = \frac{ 77 }{90}\)

That fraction is already in its lowest terms.

\(0.8\dot{ 5 } = \frac{ 77 }{ 90 }\)

The dots sit on the first and the last digit of the block that repeats, so the block is 148 and the decimal is \(0.148148148\ldots\)

Call it \(x\). Three digits repeat, so multiply by 1000:

\(1000x = 148.\dot{ 1 }4\dot{ 8 }\)

Subtracting removes the recurring tail:

\(1000x - x = 148\), which gives \(999x = 148\)

\(x = \frac{ 148 }{999}\)

The top and the bottom both divide by 37:

\(148 \div 37 = 4\) and \(999 \div 37 = 27\)

\(0.\dot{ 1 }4\dot{ 8 } = \frac{ 4 }{ 27 }\)

\(7 \div 9 = 0.777\ldots\)

One digit repeats, so it is written with a single dot: \(0.\dot{ 7 }\)

Elevenths are easiest through ninety-ninths, because \(\frac{1}{99} = 0.\dot{0}\dot{1}\).

\(\frac{ 7 }{11} = \frac{ 63 }{99}\), so the repeating block is 63.

Two digits repeat, so both the first and the last carry a dot: \(0.\dot{ 6 }\dot{ 3 }\)

Call the decimal \(x\), so \(x = 0.\dot{9}\)

One digit repeats, so multiply by 10:

\(10x = 9.\dot{9}\)

The two numbers have exactly the same digits after the point, so subtracting cancels the whole recurring tail:

\(10x - x = 9.\dot{9} - 0.\dot{9} = 9\)

\(9x = 9\), so \(x = 1\)

So \(0.\dot{9}\) is not a number just short of 1 — it is another way of writing 1, in the same way that \(0.\dot{3}\) is another way of writing \(\frac{1}{3}\) and three lots of \(\frac{1}{3}\) make exactly one whole.

The whole number part is already a whole number, so leave it alone and convert only the decimal part.

Call the decimal part \(x\), so \(x = 0.\dot{ 8 }\dot{ 2 }\)

Two digits repeat, so multiply by 100:

\(100x = 82.\dot{ 8 }\dot{ 2 }\)

\(100x - x = 82\), which gives \(99x = 82\) and \(x = \frac{ 82 }{99}\)

That fraction is already in its lowest terms.

Putting the whole number back: \(2.\dot{ 8 }\dot{ 2 } = 2\frac{ 82 }{ 99 }\)

\(2\frac{ 82 }{ 99 }\) means \(2 + \frac{ 82 }{ 99 }\).

\(2 = \frac{ 198 }{ 99 }\), so altogether

\(\frac{ 198 }{ 99 } + \frac{ 82 }{ 99 } = \frac{ 280 }{ 99 }\)

Convert each decimal to a fraction first.

\(0.6\dot{8} = \frac{ 6 }{10} + \frac{ 8 }{90} = \frac{ 62 }{90}\)

\(0.\dot{6}\dot{5} = \frac{ 65 }{99}\)

One decimal has a non-recurring digit before it starts repeating and the other does not, so the two recurring blocks do not line up. There is no shortcut here — a genuine common denominator is needed.

The lowest common denominator of 90 and 99 is 990.

\(\frac{ 62 }{90} = \frac{ 682 }{ 990 }\) and \(\frac{ 65 }{99} = \frac{ 650 }{ 990 }\)

\(\frac{ 682 }{ 990 } + \frac{ 650 }{ 990 } = \frac{ 1332 }{ 990 }\)

The top and the bottom both divide by 18:

\(1332 \div 18 = 74\) and \(990 \div 18 = 55\)

\(0.6\dot{8} + 0.\dot{6}\dot{5} = \frac{ 74 }{ 55 }\)

Call the fractional part \(y\), so \(y = 0.9\dot{1}\dot{4}\)

Since \(0.9\dot{1}\dot{4} = 0 + y\), finding \(y\) gives the printed target directly, since there is no whole number part.

One digit does not repeat, so multiply by 10 first to move the point just past it:

\(10y = 9.\dot{1}\dot{4}\)

Now multiply by 1000 to shift the point one whole block further:

\(1000y = 914.\dot{1}\dot{4}\)

Both lines have identical digits after the point, so subtracting removes the recurring tail:

\(1000y - 10y = 914 - 9\), which gives \(990y = 905\)

\(y = \frac{ 905 }{ 990 }\)

The top and the bottom both divide by 5: \(905 \div 5 = 181\) and \(990 \div 5 = 198\)

There is no whole number part, so the decimal equals \(\frac{181}{198}\), which is exactly the printed target.

Call the decimal \(x\), so \(x = 0.9\dot{6}\)

1 digit comes before the pattern starts, so multiply by 10 to move the point just past it:

\(10x = 9.\dot{6}\)

The block is 1 digit long, so multiply again by 10 to shift the point one whole block further:

\(100x = 96.\dot{6}\)

Both lines now have identical digits after the point, so subtracting removes the recurring tail:

\(100x - 10x = 96 - 9\), which gives \(90x = 87\)

\(x = \frac{ 87 }{ 90 }\)

The top and the bottom both divide by 3:

\(87 \div 3 = 29\) and \(90 \div 3 = 30\)

\(0.9\dot{6} = \frac{ 29 }{ 30 }\)

Divide 2 by 27: \(\frac{ 2 }{ 27 } = 0.\dot{0}7\dot{4}\)

The repeating block is 3 digits long, so the pattern repeats every 3 places.

The 58th decimal place is the 58th digit counting from where the block starts. Dividing 58 by 3 leaves a remainder of 1, so that place matches the 1st digit of the block.

The 1st digit of \(\dot{0}7\dot{4}\) is \(0\).