Converting Mixed Repeating Decimals

Converting Mixed Repeating Decimals | CodeSignal Learn

Introduction

0.1\overline{6}

Why Mixed Decimals Need Two Shifts

0.\overline{7}

Step by Step: Converting 0.1666...

0.1\overline{6} = 0.16666\ldots

Building Fluency: Converting 0.8333...

x = 0.8\overline{3} = 0.83333\ldots

Handling Longer Repeating Blocks

0.1\overline{72} = 0.172727272\ldots

Handling Longer Non-repeating Prefixes

0.91\overline{6} = 0.91666\ldots

The General Formula for Mixed Repeating Decimals

p

Conclusion and Next Steps

In this lesson, we extended the algebraic elimination method to mixed repeating decimals by introducing a second power of ten. The smaller shift moves past the non-repeating prefix, the larger shift moves past the prefix plus one full repeating block, and subtracting the two cancels the repeating tail cleanly. We practiced with prefixes and blocks of different lengths and distilled the approach into a general formula. Now it is time to put this technique into action! In the upcoming practice exercises, you will complete a guided conversion step by step, write full derivations from scratch, tackle mixed decimals of increasing complexity, and even convert a real-world measurement into an exact fraction. Let's turn that fresh knowledge into lasting mastery!

Previous Lesson

Join the 1M+ learners on CodeSignal

Be a part of our community of 1M+ users who develop and demonstrate their skills on CodeSignal

Welcome to the final stop in Mastery of Repeating Decimals! Over the first three lessons you built a powerful toolkit: reading bar notation, identifying repeating blocks in mixed expansions, and using the algebraic elimination method to convert pure repeating decimals into exact fractions. Now we are ready for the capstone skill — converting mixed repeating decimals into exact fractions.

A mixed repeating decimal has one or more non-repeating digits after the decimal point before the cycle begins, like $0.1\overline{6}$ or $0.91\overline{6}$ . This lesson builds directly on the four-step technique from Lesson 3. The core idea stays the same — multiply, subtract, and cancel the repeating tail — but the new wrinkle is that mixed decimals require two different powers of ten instead of one. By the end of this lesson, you will handle any mixed repeating decimal with confidence.

In the previous lesson, converting a pure repeating decimal like $0.\overline{7}$ required just one multiplication by $10$ . We multiplied, subtracted the original, and the repeating tails canceled neatly. That worked because the cycle started immediately after the decimal point, so a single shift by one block length was enough to line everything up.

Now consider $0.1\overline{6} = 0.16666\ldots$ The repeating block ( $6$ ) does not begin until after the prefix digit $1$ . If we try the old approach and multiply only by $10$ , we get $10x = 1.6666\ldots$ Subtracting the original $x = 0.1666\ldots$ does cancel the repeating part, giving $9x = 1.5$ , but the right side is a decimal rather than a whole number.

The calculation still leads to the correct answer, yet it forces an extra step of clearing that decimal. With longer prefixes or blocks, this workaround grows messy fast.

The cleaner strategy uses two multiplications: one that shifts past the prefix and one full repeating block, and another that shifts past just the prefix. Both shifted versions then share identical repeating tails starting right after the decimal point, so subtracting them produces a whole number on the right side every time.

Diagram showing the prefix and repeating block of 0.16666... with two shift arrows indicating the smaller and larger multiplications

Let's see the two-shift method in full detail with $0.1\overline{6} = 0.16666\ldots$ Here the non-repeating prefix has 1 digit and the repeating block has 1 digit.

Step 1 — Define a variable.

x = 0.16666\ldots

Step 2 — Apply the larger shift. Multiply by $10^{1+1} = 10^2 = 100$ to move past the prefix and one full block:

100x = 16.6666\ldots

Step 3 — Apply the smaller shift. Multiply by $10^1 = 10$ to move past just the prefix:

10x = 1.6666\ldots

Step 4 — Subtract. Both $100x$ and $10x$ now share the same repeating tail $.6666\ldots$ after the decimal point:

100x - 10x = 16.6666\ldots - 1.6666\ldots

90x = 15

Vertical alignment diagram showing 100x minus 10x with repeating tails highlighted and aligned to show cancellation, yielding 90x equals 15

Step 5 — Solve and simplify.

x = \frac{15}{90} = \frac{1}{6}

We divided numerator and denominator by $15$ . So $0.1\overline{6} = \frac{1}{6}$ . A quick check confirms this: $1 \div 6 = 0.1666\ldots$ ✓

Let's try another example with the same structure — one prefix digit and a one-digit block — and move through it a bit more briskly.

Let $x = 0.8\overline{3} = 0.83333\ldots$ The prefix has $1$ digit and the block has $1$ digit, so we multiply by $100$ and by $10$ :

100x = 83.3333\ldots

10x = 8.3333\ldots

Subtract:

90x = 75

x = \frac{75}{90} = \frac{5}{6}

We simplified by dividing by $15$ . Notice the denominator before simplification was $90$ in both this example and the previous one. That is not a coincidence — whenever the prefix is $1$ digit and the block is $1$ digit, we always get $10^2 - 10^1 = 90$ .

When the repeating block has more than one digit, the larger power of ten increases to match. Let's convert $0.1\overline{72} = 0.172727272\ldots$ , which has a 1-digit prefix and a 2-digit repeating block.

Larger shift: $10^{1+2} = 10^3 = 1{,}000$
Smaller shift: $10^1 = 10$

1000x = 172.7272\ldots

10x = 1.7272\ldots

Subtract:

990x = 171

x = \frac{171}{990} = \frac{19}{110}

We divided by $9$ since $\gcd(171,\, 990) = 9$ . The denominator before simplifying was $990 = 1{,}000 - 10$ , which fits the emerging pattern: the denominator is always $10^{p+r} - 10^p$ , where $p$ is the number of prefix digits and $r$ is the length of the repeating block.

What happens when the prefix is longer than one digit? The same logic applies — we simply increase the smaller power of ten. Let's convert $0.91\overline{6} = 0.91666\ldots$ , which has a 2-digit prefix ( $91$ ) and a 1-digit repeating block ( $6$ ).

Larger shift: $10^{2+1} = 10^3 = 1{,}000$
Smaller shift: $10^2 = 100$

1000x = 916.666\ldots

100x = 91.666\ldots

Subtract:

900x = 825

x = \frac{825}{900} = \frac{11}{12}

We divided by $75$ . The key takeaway here is that the smaller power of ten always matches the number of non-repeating digits. This ensures both shifted versions have the repeating part starting right at the decimal point, so the tails cancel perfectly.

Here is a nice real-world connection: if a stopwatch reads $0.91\overline{6}$ hours, that is exactly $\frac{11}{12}$ of an hour, or $55$ minutes. Fractions like $\frac{11}{12}$ are far easier to interpret and reuse in further calculations than an infinite decimal string.

Let's collect everything into a tidy summary. For a mixed repeating decimal with $p$ non-repeating digits after the decimal point and $r$ repeating digits:

Component	Value
Larger shift	$10^{p+r}$
Smaller shift	$10^{p}$
Denominator (before simplifying)	$10^{p+r} - 10^{p}$

The subtraction on the right side always yields a whole number equal to the digits from the decimal point through one full block, minus the prefix digits alone. Written as a formula:

x = \frac{(\text{prefix + one block as a number}) - (\text{prefix as a number})}{10^{p+r} - 10^{p}}

For example, with $0.1\overline{72}$ : the numerator is $172 - 1 = 171$ and the denominator is $1{,}000 - 10 = 990$ , giving $\frac{171}{990}$ before simplification. As always, the final step is to divide numerator and denominator by their greatest common factor to reach simplest form.

Conceptual diagram of a generic mixed repeating decimal with p non-repeating and r repeating digits labeled, showing the corresponding shifts and denominator formula