GMAT® Terminating and Repeating Decimals: A Complete Guide

Dividing out a fraction to get a decimal is comfortable work. You've done it since school, the basic steps stay the same, and a question that hands you $\frac{7}{20}$ and asks for a decimal can look like a gift: one round of division and the answer sits there.

But some fractions refuse to cooperate. Divide 1 by 3 and you get $0.333...$, and the 3s never stop no matter how long you keep dividing. Questions built on that fact rarely ask you to finish the division. They ask which kind of decimal you're in for before you start, and that answer is readable straight off the denominator once you've seen the method.

Two kinds of decimals

Any fraction made of whole numbers produces a decimal of exactly one of two kinds. A terminating decimal stops. It has a last digit, the division ends cleanly, and $\frac{1}{4} = 0.25$ is the everyday version. A repeating decimal falls into a cycle instead: $\frac{1}{3} = 0.333...$, with the 3 continuing forever.

The standard notation for that second kind is a bar over the repeating block, so $0.333...$ gets written $0.\overline{3}$. The bar means "this block repeats without end," which lets one symbol stand in for a number that never got around to ending.

There's also a mixed kind. $\frac{1}{6}$ comes out $0.1666...$: a 1, then 6s forever. Only part of the decimal cycles, so it gets written $0.1\overline{6}$. It repeats and never terminates, but the repeating part doesn't start right away, and questions lean on that gap.

Long division decides

Every fraction's decimal comes from one process: divide, multiply, subtract, bring down the next digit. Our Math Basics series calls that cycle DMSB and walks through the mechanics in detail, so this section sticks to what the cycle reveals.

Watch the remainders in a division of $\frac{1}{6}$:

    0.1 6 6
    ---------
6 ) 1.0 0 0 0
    0                6 into 1 goes 0 times
    -
    1 0              bring down the first 0: 10 tenths
      6              6 into 10 goes 1 time
      -
      4              remainder 4
    ---
      4 0            bring down: 40 hundredths
      3 6            6 into 40 goes 6 times
    ---
        4            remainder 4 again
    ---
        4 0          the same remainder returns, so the 6s repeat

Each digit of the answer is decided by a remainder, and a division by 6 can only ever produce six different remainders: 0 through 5. When the remainder 4 showed up a second time, everything downstream had to happen again in the same order, which is why the 6s repeat. A remainder of 0 is the other exit, and it means the division came out even, the decimal stops, and every digit after would be another 0.

That leaves no third possibility. If the remainder never hits 0, it has to run out of new values and revisit one, and the moment it does, the cycle locks in. Every fraction of whole numbers terminates or repeats, and long division is where you can watch one of the two happen. The cycle has to appear within as many steps as the divisor has possible remainders, so a division by 12 settles its own case within twelve cycles, usually well before that.

The reduce-first test

Long division settles the terminate-or-repeat question for any fraction of whole numbers, but you rarely need it. The denominator alone can answer it, with one preliminary move: reduce the fraction first.

A fraction in lowest terms terminates when its denominator's prime factors are only 2s and 5s, and repeats when anything else is in there. So $\frac{7}{20}$ terminates, because $20 = 2^2 \times 5$, while $\frac{5}{12}$ repeats, because $12 = 2^2 \times 3$ and that 3 has nowhere to go.

The 2s and 5s rule comes straight from what a terminating decimal is. A decimal that stops is a number of tenths, or hundredths, or thousandths: some count of a power of 10. $\frac{1}{8}$ terminates because you can scale it to thousandths by multiplying top and bottom by 125:

$\frac{1}{8} = \frac{125}{1000} = 0.125$

Powers of 10 are built from 2s and 5s alone, since $10^n = 2^n \times 5^n$. A reduced denominator made of only 2s and 5s can be scaled up into some power of 10, which means the decimal lands on a count of tenths, hundredths, or thousandths and stops. A denominator carrying any other prime can never divide a power of 10, so the division never settles and the remainders cycle instead.

The reduce-first part is where the trap lives, because the original denominator can mislead you. $\frac{3}{12}$ looks like it repeats: 12 carries a 3. Reduce it and you get $\frac{1}{4} = 0.25$, which terminates, because the 3 was the numerator's problem and it canceled away. The primes that matter are the ones still standing after reducing, so cancel everything that cancels, then read what's left.

Two more quick reads to calibrate:

$\frac{7}{20} = 0.35 \qquad \frac{9}{40} = 0.225$

$20 = 2^2 \times 5$ and $40 = 2^3 \times 5$, so both terminate, and neither division needs to be run to know it. When you do want the decimal itself, the reduced denominator tells you how many places it takes: $2^a \times 5^b$ terminates in exactly $\max(a, b)$ decimal places.

Where the GMAT® uses this

The direct version shows up in Data Sufficiency, where "does $\frac{a}{b}$ have a terminating decimal representation" is a question about the reduced denominator's prime factors. A statement that $\frac{a}{b}$ terminates constrains the reduced form, not the original numbers. $b$ itself can carry a 3, as $\frac{3}{12}$ shows, because the numerator can cancel it. Our Data Sufficiency guide covers how statements like this slot into the broader framework.

The harder version asks about a digit far past where anyone would divide. Take $\frac{1}{7} = 0.\overline{142857}$, a cycle of six digits. What's the 50th digit after the decimal point? Dividing out 50 places isn't practical, but the cycle answers it: $50 = 8 \times 6 + 2$, so the 50th digit sits where the 2nd digit of the cycle sits, which is 4. Long division builds the cycle once, and from there it covers every position a question can ask for.

The rounding connection runs the other way. Our decimal rounding guide recommends dividing one place past what the question needs, and this topic is part of why that advice works: quotients on this exam often don't terminate, so the extra place is how you find the deciding digit in a decimal that was never going to stop.

Try this one

Try this one before reading on.

Reduce each choice first, then read the denominator's primes. (A) has 11 in the denominator, so it repeats. (B) keeps $2^2 \times 3$, so it repeats. (C) keeps $2^3 \times 3$, so it repeats. (D) is already in lowest terms with $40 = 2^3 \times 5$, only 2s and 5s, so it terminates: $\frac{9}{40} = 0.225$. (E) keeps $3 \times 5$, so it repeats. The answer is (D).

FAQ

What is a terminating decimal on the GMAT®?

A terminating decimal is one that stops, like $0.25$ or $0.225$. A fraction in lowest terms terminates when its denominator's only prime factors are 2 and 5, because such a denominator can be scaled into a power of 10.

What is a repeating decimal?

A repeating decimal falls into a cycle of digits that continues forever, like $\frac{1}{3} = 0.\overline{3}$. The bar over the digits marks the repeating block. Any fraction whose reduced denominator contains a prime other than 2 or 5 repeats.

How do you tell whether a fraction terminates or repeats?

Reduce the fraction completely, then factor the denominator. If only 2s and 5s remain, the decimal terminates. If any other prime remains, the decimal repeats. Long division confirms it: a remainder of 0 means terminate, and a repeated remainder means the cycle has begun.

Why do only 2s and 5s in the denominator terminate?

Because powers of 10 are made of only 2s and 5s: $10^n = 2^n \times 5^n$. A terminating decimal is a count of tenths, hundredths, or thousandths, so its denominator has to divide some power of 10. Denominators built from 2s and 5s can; denominators carrying any other prime can't.

Does 1/6 terminate or repeat on the GMAT®?

It repeats, but only part of it: $\frac{1}{6} = 0.1\overline{6}$. The reduced denominator 6 contains a 3, so the decimal can't terminate, and the division shows the tail: the remainder 4 returns every cycle and the 6 repeats forever.

How do you find a digit deep inside a repeating decimal?

Divide just far enough to establish the repeating block, then count positions within the cycle. For $\frac{1}{7} = 0.\overline{142857}$, the cycle runs six digits, and 50 divided by 6 leaves a remainder of 2, so the 50th digit is the 2nd digit of the cycle, which is 4.

Want to learn even more?

Related reading:

• GMAT® Number Properties: A Complete Guide
• GMAT® Divisibility Rules: A Complete Guide
• GMAT® Least Common Multiples: A Complete Guide
• GMAT® Decimal Rounding Rules: A Complete Guide
• GMAT® Data Sufficiency: How to Approach DS Questions

The Math Basics series rebuilds decimal arithmetic from the ground up, including "GMAT® Focus Edition Math Basics: Decimal Long Division," which runs the DMSB cycle step by step, and "GMAT® Focus Edition Math Basics: Division with Decimals," which extends the same cycle to divisors with decimal points of their own. You can hear both on Spotify, Apple Podcasts, and YouTube.