See a Repeating Decimal in a Problem? Deploy the 100x Minus x Trick

A JEE problem hands you 0.2\overline{37} and says "express as a rational p/q." A board question shows 0.\overline{142857} and asks for its fraction form. In every case, the exam is testing the same instinct: see the repeating decimal, reach for the multiply-and-subtract trick. The calculation is always the same three lines. The only thing that changes is the power of 10 you multiply by.

The recognition trigger

Any of these phrases, in any order, should fire the same move in your head.

"Express the following decimal as a rational number."
"Convert the repeating decimal to p/q form."
"Write in the form p/q where p, q are integers."
"Show that 0.\overline{ab} is rational."

The answer format is always a fraction. The input is always a decimal with a repeating tail. The method is always the same: multiply x by 10^k where k is the length of the repeat block, subtract the original x, solve.

The move, in three lines

Let x be the decimal. Multiply by the power of 10 that shifts the repeat block by exactly one period. Subtract. The repeating tail cancels, leaving a linear equation in x.

For x = 0.\overline{7} (period 1):

10x = 7.\overline{7}, \qquad 10x - x = 7, \qquad 9x = 7, \qquad x = \tfrac{7}{9}.

For x = 0.\overline{37} (period 2):

100x = 37.\overline{37}, \qquad 100x - x = 37, \qquad 99x = 37, \qquad x = \tfrac{37}{99}.

For x = 0.\overline{142857} (period 6):

10^6 x = 142857.\overline{142857}, \qquad 10^6 x - x = 142857, \qquad 999999 x = 142857, \qquad x = \tfrac{142857}{999999} = \tfrac{1}{7}.

Notice the pattern on the right-hand side: the denominator is always a string of 9s with length equal to the period. A 1-digit block goes over 9; a 2-digit block goes over 99; a 6-digit block goes over 999999.

Three different repeating decimals, three different period lengths, one universal rule: the block over a string of $9$s of the same length. Simplify at the end (the $142857$ case reduces to $1/7$).

When the repeating starts after a non-repeating prefix

Sometimes the decimal has a non-repeating part before the repeat kicks in — like 0.2\overline{37}. The same move still works; you just multiply by two powers of 10 and subtract.

Let x = 0.2\overline{37}. Shift past the non-repeating prefix (10x = 2.\overline{37}), then shift by the period (1000x = 237.\overline{37}). Subtract:

1000x - 10x = 237 - 2 \implies 990x = 235 \implies x = \tfrac{235}{990} = \tfrac{47}{198}.

The general rule: if the decimal has m non-repeating digits followed by a k-digit repeat, multiply by 10^m and by 10^{m+k}, subtract, solve. Alternatively, write the fraction straight from memory: the denominator is k nines followed by m zeros; the numerator is (non-repeating digits followed by one period) minus (non-repeating digits alone).

Check with 0.2\overline{37}: one non-repeating digit (m = 1), two-digit repeat (k = 2). Denominator is 99 followed by one zero: 990. Numerator is 237 - 2 = 235. Fraction: \tfrac{235}{990} = \tfrac{47}{198}. Matches.

Why multiplying by 10^k kills the tail

The repeating decimal is a number whose tail, after a certain point, is periodic. Multiplying by 10^k shifts the entire decimal expansion left by k places — and because the tail has period k, the shifted tail is identical to the original tail. When you subtract the original, the tails wipe each other out, leaving only the shifted "head" minus the original "head" — an integer difference on the right, and a finite multiple of x on the left.

Why the cancellation is exact, not approximate: if x = 0.d_1 d_2 \ldots d_k d_1 d_2 \ldots d_k d_1 \ldots, then 10^k x shifts every digit left by k places. Because the block d_1 d_2 \ldots d_k repeats starting from position 1, the tail of 10^k x starting at position 1 is the exact same sequence of digits as the tail of x starting at position 1. Subtraction cancels them digit-for-digit, leaving a finite integer on the right.

The sanity checks that save marks

Three quick ways to sanity-check a repeating-decimal-to-fraction conversion.

Denominator of 9s of the right length. Before simplification, your denominator should be 99 \ldots 9 with the repeat length. If it's not, you counted the period wrong.
The fraction simplified correctly. \tfrac{142857}{999999} is not in lowest terms; it reduces to \tfrac{1}{7}. The graded answer often expects lowest terms. Always reduce.
Convert back to verify. Take your final fraction and divide by hand for one or two digits. \tfrac{47}{198}: 47 / 198 \approx 0.237\ldots — looks right. \tfrac{37}{99}: 37/99 \approx 0.3737\ldots — checks out.

If any of these fails, recount the period or the non-repeating digits.

Convert $0.7\overline{21}$ to a fraction

x = 0.7\overline{21}. Non-repeating prefix: one digit (7). Repeat block: two digits (21).

Shift past the prefix: 10x = 7.\overline{21}. Shift past one full period: 1000x = 721.\overline{21}. Subtract: 1000x - 10x = 721 - 7 = 714. So 990 x = 714, giving x = \tfrac{714}{990}.

Reduce: \gcd(714, 990) = 6, so x = \tfrac{119}{165}.

Sanity check: 119/165 \approx 0.7212\ldots — matches the original.

Why this is faster than memorising a big formula: you only need to remember two shifts (10^m past the prefix, 10^{m+k} past prefix-plus-one-period) and the subtraction. Everything else falls out of the arithmetic.

What to remember

Problems asking "express as a rational" or "convert to p/q" with a repeating decimal → this trick, every time.
Multiply x by 10^k where k is the period. Subtract the original. The tail cancels. Solve for x.
If there's a non-repeating prefix of length m, use two multiplications (10^m and 10^{m+k}), subtract, solve.
The denominator before simplification is always \underbrace{99\ldots 9}_{k}\underbrace{00\ldots 0}_{m} — a memorable pattern that doubles as a sanity check.
Reduce to lowest terms before writing the final answer.

The move is so mechanical you should not have to think about it; the instant your eyes see "decimal with a bar" and "rational" in the same sentence, the fingers should already be writing "Let x = \ldots." That is the recognition.