Spot a Terminating Fraction by the Denominator — 2s and 5s Only, or It Repeats

When a problem shows you a fraction and asks whether its decimal terminates or repeats, the reflex most students use is "divide it out and see." That works, but it is slow, and on a JEE paper you need the answer before the long division is finished. Here is the recognition shortcut that settles it in the time it takes to read the denominator.

The rule, in one sentence

Reduce the fraction to lowest terms. Factorise the denominator into primes. If the only primes are 2 and 5, the decimal terminates. Any other prime — even one — forces the decimal to repeat.

That is the whole rule. Say it out loud once, and the check takes five seconds on any fraction you see.

Applying it on sight

Read a fraction, look only at the denominator, factor it, check for 2s and 5s.

\tfrac{7}{40}. Denominator 40 = 2^3 \cdot 5. Only 2s and 5s. Terminates. (Verify: 0.175.)
\tfrac{1}{3}. Denominator 3. Not 2 or 5. Repeats. (Verify: 0.\overline{3}.)
\tfrac{9}{16}. Denominator 16 = 2^4. Only 2s. Terminates. (Verify: 0.5625.)
\tfrac{5}{12}. Denominator 12 = 2^2 \cdot 3. The 3 breaks the rule. Repeats. (Verify: 0.41\overline{6}.)
\tfrac{17}{250}. Denominator 250 = 2 \cdot 5^3. Only 2s and 5s. Terminates. (Verify: 0.068.)
\tfrac{3}{14}. Denominator 14 = 2 \cdot 7. The 7 breaks the rule. Repeats. (Verify: 0.2142857\overline{142857}.)

The pattern is visual. You are scanning for "does any prime other than 2 or 5 appear?" The answer takes the same time as reading the fraction.

Six fractions sorted by the two-primes test. Left: denominators built only from $2$ and $5$. Right: denominators with at least one prime from outside $\{2, 5\}$. The decimal behaviour follows the column.

Why it works in half a line

A terminating decimal is a fraction whose denominator is a power of 10: 0.175 = \tfrac{175}{1000} = \tfrac{175}{10^3}. Since 10 = 2 \cdot 5, a power of 10 is built entirely from 2s and 5s. So a fraction terminates exactly when you can rewrite it with a power of 10 on the bottom — and that is possible exactly when the denominator already has only 2s and 5s (you can then balance the counts by multiplying top and bottom by whatever is missing).

Why: \tfrac{7}{40} = \tfrac{7}{2^3 \cdot 5} = \tfrac{7 \cdot 5^2}{2^3 \cdot 5^3} = \tfrac{175}{1000} = 0.175. The multiplication by 5^2 is legal because you are multiplying top and bottom by the same thing. This trick works exactly when the original denominator has only 2s and 5s. Any other prime — say 3 — would need a factor of 3 in 10^k too, but 10^k never has a 3, so the balancing is impossible and the fraction cannot be put over a power of ten.

The "in lowest terms" catch

The test only works on a fraction in lowest terms. If you forget to reduce first, you will get the wrong answer.

\tfrac{3}{6} has denominator 6 = 2 \cdot 3. The 3 seems to predict repetition. But \tfrac{3}{6} = \tfrac{1}{2}, which is 0.5 — terminating. The 3 on top cancelled the 3 on the bottom, and the true denominator is just 2.

So your first move, always, is to reduce. \tfrac{15}{24} = \tfrac{5}{8}, denominator 8 = 2^3, terminates. \tfrac{21}{70} = \tfrac{3}{10}, denominator 10 = 2 \cdot 5, terminates. \tfrac{22}{33} = \tfrac{2}{3}, denominator 3, repeats.

Reduce first, factor second, judge third.

What this recognition buys you in exam time

In MCQ land, this test answers "which of these fractions has a terminating decimal?" without any division. Sample question:

Which of these fractions has a terminating decimal expansion?
(a) \tfrac{7}{45} (b) \tfrac{9}{200} (c) \tfrac{11}{60} (d) \tfrac{13}{70}

Scan the denominators after reducing:

45 = 3^2 \cdot 5. Has 3. Repeats.
200 = 2^3 \cdot 5^2. Only 2s and 5s. Terminates.
60 = 2^2 \cdot 3 \cdot 5. Has 3. Repeats.
70 = 2 \cdot 5 \cdot 7. Has 7. Repeats.

Answer: (b). Time: about ten seconds. No long division involved.

Sort these six at a glance

\frac{3}{8}, \quad \frac{4}{15}, \quad \frac{7}{125}, \quad \frac{11}{6}, \quad \frac{9}{50}, \quad \frac{13}{22}.

\tfrac{3}{8}: denominator 8 = 2^3. Terminates.
\tfrac{4}{15}: denominator 15 = 3 \cdot 5. Has 3. Repeats.
\tfrac{7}{125}: denominator 125 = 5^3. Terminates.
\tfrac{11}{6}: denominator 6 = 2 \cdot 3. Has 3. Repeats.
\tfrac{9}{50}: denominator 50 = 2 \cdot 5^2. Terminates.
\tfrac{13}{22}: denominator 22 = 2 \cdot 11. Has 11. Repeats.

Three terminate, three repeat. Total time on the recognition: under twenty seconds for all six. Zero long divisions.

Why it's so fast: factorising small denominators is a grade-school reflex, and scanning a prime factorisation for "anything other than 2 or 5" is a visual operation — you do not need to reason, only to look.

What to remember

In lowest terms, a fraction's decimal terminates iff the denominator has only 2s and 5s as prime factors.
Reduce before factoring — an uncancelled common factor can change which column the fraction falls into.
The reason is that a terminating decimal is exactly a fraction over a power of 10, and powers of 10 are built only from 2s and 5s.
The check takes about five seconds per fraction, making it the standard tool for MCQ questions that ask "which terminates?"

So "denominator has only 2s and 5s" is not a random fact you memorise — it is the only way the denominator can match a power of ten, which is the only way a decimal can end. Every other fraction repeats, and it does so because no amount of climbing up powers of 10 will ever catch up with a 3 or a 7 lurking in the bottom.