Repeating Decimals

Bring down a zero and divide. \(10 \div 3\) gives digit \(3\) with \(10 - 9 = 1\) left over. Bring down another zero and \(10 \div 3\) gives \(3\) again with \(1\) left over. The remainder at the end of every step is \(\boxed{1}\). Since that leftover is the number we started with, the step copies itself forever, which is why \(\tfrac{1}{3} = 0.\overline{3}\).

Follow the remainders. \(10 \div 7\) gives \(1\) remainder \(3\), then \(30 \div 7\) gives \(4\) remainder \(2\), then \(20 \div 7\) gives \(2\) remainder \(6\), then \(60 \div 7\) gives \(8\) remainder \(4\), then \(40 \div 7\) gives \(5\) remainder \(5\), then \(50 \div 7\) gives \(7\) remainder \(1\). The starting remainder is back after six steps, so the block is \(\boxed{6}\) digits long, namely \(142857\). A block can never run longer than the count of available nonzero remainders, and here all six of \(1\) through \(6\) get used.

Divide \(1\) by \(6\). \(10 \div 6 = 1\) remainder \(4\), so the first digit is \(1\). Then \(40 \div 6 = 6\) remainder \(4\), the same remainder, so the \(6\) repeats from there. That gives \(\tfrac{1}{6} = 0.1\overline{6}\), and only \(\boxed{1}\) digit sits outside the bar. The head comes from the \(2\) in \(6 = 2 \times 3\). One factor of \(2\) buys one settling digit, and the \(3\) makes the loop.

\(10 \div 11 = 0\) remainder \(10\), so the first digit after the point is \(0\). Then \(100 \div 11 = 9\) remainder \(1\), which is the remainder we began with, so the digits cycle as \(0, 9, 0, 9, \ldots\) and \(\tfrac{1}{11} = 0.\overline{09}\). The block \(09\) has \(\boxed{2}\) digits. Keep that leading zero, since \(0.\overline{9}\) is a completely different number.

Divide \(1\) by \(12\). \(10 \div 12 = 0\) remainder \(10\), then \(100 \div 12 = 8\) remainder \(4\), then \(40 \div 12 = 3\) remainder \(4\). Remainder \(4\) has come back, so that step repeats forever and \(\tfrac{1}{12} = 0.08\overline{3}\). The block under the bar is \(\boxed{3}\). The head \(08\) is two digits because \(12 = 2^2 \times 3\) carries two factors of \(2\), and the \(3\) is what loops.

Factor the four denominators. \(16 = 2^4\), \(20 = 2^2 \times 5\), \(25 = 5^2\), and \(15 = 3 \times 5\). The first three are built only from \(2\)s and \(5\)s, so they terminate at \(0.0625\), \(0.05\), and \(0.04\). Only \(15\) carries a \(3\), and \(\tfrac{1}{15} = 0.0\overline{6}\), so the denominator is \(\boxed{15}\). Only \(2\) and \(5\) divide a power of ten, so any other prime can never be scaled away and the remainders cycle instead of reaching \(0\).

Let \(x = 0.7777\ldots\), so \(10x = 7.7777\ldots\). Past the decimal point both numbers carry the identical endless string of \(7\)s, so subtracting wipes it out. $$10x - x = 7.7777\ldots - 0.7777\ldots$$ $$9x = 7$$ Divide by \(9\) to get \(x = \tfrac{7}{9}\), and \(7\) is prime so nothing reduces. $$\boxed{7/9}$$ Multiplying by \(10\) shifts past exactly one block, which is what lines the two tails up.

Let \(x = 0.121212\ldots\). The block is two digits, so multiply by \(100\). $$100x = 12.121212\ldots$$ The tails match, so subtracting leaves \(99x = 12\) and \(x = \tfrac{12}{99}\). Both parts are divisible by \(3\). $$\frac{12}{99} = \frac{4}{33}$$ Since \(4 = 2 \times 2\) and \(33 = 3 \times 11\), this is fully reduced, so \(0.\overline{12} = \boxed{4/33}\). The block-over-nines shortcut hands you a correct fraction but rarely a reduced one, so always check.

Let \(x = 0.9999\ldots\), so \(10x = 9.9999\ldots\). After the decimal point both numbers carry the same endless string of nines, so subtracting cancels it. $$10x - x = 9.9999\ldots - 0.9999\ldots$$ $$9x = 9$$ Divide by \(9\) and \(x = \boxed{1}\). Nothing was rounded or dropped anywhere in that work, so \(0.\overline{9}\) is not creeping up toward \(1\). It is another name for \(1\).

Let \(x = 0.41666\ldots\). Shift past the head with \(100x = 41.666\ldots\), then one block further with \(1{,}000x = 416.666\ldots\). Both end in the same \(.666\ldots\) tail, so subtracting cancels it. $$1{,}000x - 100x = 416.666\ldots - 41.666\ldots$$ $$900x = 375$$ Then \(x = \tfrac{375}{900}\), and both share a factor of \(75\), since \(375 = 75 \cdot 5\) and \(900 = 75 \cdot 12\). $$x = \boxed{5/12}$$ When there is a non-repeating head you need two shifts, one landing just before the loop and one a full block later.

The block \(142857\) is six digits long, so divide. \(50 = 6 \times 8 + 2\), which puts position \(50\) two digits into a fresh block. The second digit of \(142857\) is \(\boxed{4}\). Only the remainder picks the digit. The quotient \(8\) just counts how many full blocks you skipped past.

A two-digit block sits over \(99\), so \(0.\overline{72} = \tfrac{72}{99}\) and \(0.\overline{27} = \tfrac{27}{99}\). The denominators already match, so subtract the numerators. $$\frac{72}{99} - \frac{27}{99} = \frac{45}{99}$$ Both \(45\) and \(99\) are divisible by \(9\), which gives \(\tfrac{5}{11}\), and \(5\) shares no factor with \(11\). $$\boxed{5/11}$$ Checking, \(\tfrac{5}{11} = 0.\overline{45}\), and \(0.7272\ldots\) minus \(0.2727\ldots\) does leave \(0.4545\ldots\).

Since \(3\) is bigger than \(2\), the whole-number part is \(0\) and the remainder is \(2\). Then \(20 \div 3 = 6\) with remainder \(20 - 18 = 2\), the remainder we started with, so every step after that prints another \(6\). That makes \(2 \div 3 = 0.\overline{6}\), so the repeating digit is \(\boxed{6}\). A returning remainder is exactly what locks a decimal into a loop.

Bring down a zero to get \(10\). Then \(10 \div 9 = 1\) with remainder \(1\), the number we started with, so every step writes another \(1\) and \(\tfrac{1}{9} = 0.\overline{1}\). The bar covers a single digit, so the period is \(\boxed{1}\).

Start with remainder \(1\). Attach a zero to get \(10\), and \(11\) fits into \(10\) zero times, so the first digit is \(0\) and the remainder is \(10\). Attach another zero to get \(100\), and \(11\) fits \(9\) times since \(9 \times 11 = 99\), leaving remainder \(1\) again. The digits cycle as \(0, 9, 0, 9, \ldots\), so the block is \(\boxed{09}\). The leading zero is part of the block, not decoration.

\(50 \div 6 = 8\) with remainder \(2\), so the first decimal digit is \(8\). Then \(20 \div 6 = 3\) with remainder \(2\) again, so the \(3\) loops from there and \(\tfrac{5}{6} = 0.8\overline{3}\). Exactly \(\boxed{1}\) digit sits outside the bar. A decimal like this is called eventually repeating, since it settles into its loop after a short run-up instead of right away.

\(70 \div 30 = 2\) with remainder \(10\), so the first decimal digit is \(2\). Then \(100 \div 30 = 3\) with remainder \(10\), the same remainder, so every later step writes another \(3\). That gives \(\tfrac{7}{30} = 0.2\overline{3}\), so the repeating block is \(\boxed{3}\). The \(2\) and the \(5\) in \(30\) produce the one settling digit, and the \(3\) produces the loop.

Let \(x = 0.5555\ldots\), so \(10x = 5.5555\ldots\). Both numbers end in the identical endless string of fives, so subtracting cancels the tails and leaves \(9x = 5\). Divide by \(9\) to get \(x = \tfrac{5}{9}\), and since \(9 = 3 \times 3\) shares nothing with \(5\), it is already in lowest terms. \(\boxed{5/9}\)

The block is one digit, so it goes over a single nine, and \(0.\overline{2} = \tfrac{2}{9}\). The factors of \(2\) are \(1\) and \(2\), the factors of \(9\) are \(1\), \(3\), and \(9\), so they share only \(1\) and the fraction cannot shrink. \(\boxed{2/9}\) That nine is really \(10 - 1\). Setting \(x = 0.\overline{2}\) and subtracting gives \(10x - x = 9x = 2\).

A two-digit block sits over \(99\), so \(0.\overline{81} = \tfrac{81}{99}\). The greatest common factor of \(81\) and \(99\) is \(9\), since \(81 = 9 \times 9\) and \(99 = 9 \times 11\). Dividing top and bottom by \(9\) gives \(\boxed{9/11}\). Checking, \(9 \div 11 = 0.818181\ldots\). The block-over-nines shortcut gives a correct value but rarely a reduced one.

Let \(x = 0.0454545\ldots\). Shift three places to land just before a fresh block, \(1{,}000x = 45.4545\ldots\), and one place to clear the head zero, \(10x = 0.4545\ldots\). Both end in the same \(.4545\ldots\) tail, so subtracting gives \(990x = 45\) and \(x = \tfrac{45}{990}\). Both share \(45\), since \(990 = 45 \cdot 22\), so \(x = \boxed{1/22}\). The leading zero appears once and never repeats, which is why a single shift will not line the tails up.

Let \(x = 0.\overline{9}\), so \(10x = 9.\overline{9}\). After the decimal point both carry the identical endless string of nines, so subtracting cancels it and leaves \(9x = 9\). Divide by \(9\) and \(x = \boxed{1}\). A second check agrees. \(0.\overline{1} = \tfrac{1}{9}\), and nine copies give \(\tfrac{9}{9} = 1\) while \(9 \cdot 0.\overline{1} = 0.\overline{9}\).

The block \(076923\) is six digits long, so divide. \(100 = 6 \times 16 + 4\), since \(6 \times 16 = 96\), which puts position \(100\) four digits into a fresh block. Reading \(076923\), the 4th digit is \(\boxed{9}\). The quotient \(16\) only counts the full blocks that went by. The remainder is what picks the digit.

The block \(12\) is two digits long, so divide. \(99 = 2 \times 49 + 1\), so position \(99\) lands on the first digit of the block, which is \(\boxed{1}\). With a block of length \(2\), odd positions all hold the \(1\) and even positions all hold the \(2\), and \(99\) is odd.

Read the factorizations. \(8 = 2^3\), \(40 = 2^3 \times 5\), and \(50 = 2 \times 5^2\) use only \(2\)s and \(5\)s, so all three terminate. \(24 = 2^3 \times 3\) carries a \(3\), which is neither \(2\) nor \(5\), so the repeating denominator is \(\boxed{24}\). Its decimal is \(\tfrac{1}{24} = 0.041\overline{6}\), and that stray \(3\) is the reason it never ends.

A two-digit block sits over \(99\), so \(0.\overline{81} = \tfrac{81}{99}\) and \(0.\overline{36} = \tfrac{36}{99}\). The denominators match, so subtract the numerators. $$\frac{81}{99} - \frac{36}{99} = \frac{45}{99}$$ Both \(45\) and \(99\) are divisible by \(9\), since \(45 = 9 \times 5\) and \(99 = 9 \times 11\), so the answer is \(\boxed{5/11}\). Checking, \(5 \div 11 = 0.\overline{45}\), exactly the gap between the two decimals.