There is a specific phrasing that appears in number theory, algebra, and even geometry problems, and every time it appears the answer follows from one concept: the least common multiple. Memorising that mapping — phrase → concept — is the single best trade you can make. Problems that would take twenty minutes of trial and error collapse to one line.

The phrasings all look something like:

All of these are LCM problems wearing different costumes. The moment you see the structural signature "smallest positive integer that is a common multiple of this list," the work to do is:

\operatorname{lcm}(a_1, a_2, \ldots, a_k).

Why "smallest divisible by all" = LCM, precisely

The LCM of positive integers a_1, a_2, \ldots, a_k is, by definition, the smallest positive integer that is a multiple of every a_i. So:

The third point is the reason the LCM is so powerful. It says that the set of all common multiples of \{a_1, \ldots, a_k\} is exactly the set of positive multiples of \operatorname{lcm}(a_1, \ldots, a_k). So if a problem asks for "some positive integer divisible by all these," the answer is "any positive multiple of the LCM" — and if it asks for the smallest, you give the LCM itself.

Recognising the signature in disguise

Many problems do not use the word "LCM" anywhere. Here are four patterns that hide it.

Pattern 1: "When do events next coincide?"

Three traffic signals on Marine Drive turn red every 40, 48, and 60 seconds respectively. If they turn red together at 10{:}00{:}00, when will they next turn red together?

Each signal turns red at times that are multiples of its period. They coincide at times that are multiples of all three periods — that is, common multiples of 40, 48, 60. The next coincidence after 0 is the LCM.

40 = 2^3 \cdot 5, \quad 48 = 2^4 \cdot 3, \quad 60 = 2^2 \cdot 3 \cdot 5.

Take max exponents: 2^4 \cdot 3 \cdot 5 = 240. They next turn red together 240 seconds later, at 10{:}04{:}00.

Pattern 2: "Fewest objects that can be split these ways"

A school has students that can be divided evenly into teams of 5, 6, or 8. What is the smallest number of students for which this is possible?

"Divided evenly" means divisible. Smallest number divisible by 5, 6, 8 → LCM.

\operatorname{lcm}(5, 6, 8) = \operatorname{lcm}(5, 6, 2^3) = 2^3 \cdot 3 \cdot 5 = 120.

Pattern 3: "Smallest common cycle length"

Two gears mesh. Gear A has 15 teeth, gear B has 18. How many rotations of A before both gears return to their starting position?

After n rotations of A, a total of 15n teeth have passed. Gear B returns to start when 15n is a multiple of 18 — i.e., when 15n is a common multiple of 15 and 18. The smallest such product is \operatorname{lcm}(15, 18) = 90. So n = 90 / 15 = 6 rotations of A.

Pattern 4: "Combining periods"

An event repeats every 3 days, another every 5 days, a third every 7 days. If all occur today, when is the next day all three occur?

LCM of 3, 5, 7 = 105 days (since they are pairwise coprime, LCM is the product).

In every one of these, no one used the word "LCM" — but the structural cue "smallest number divisible by all of these" or its equivalent "first time all cycles coincide" is the clue. Once you see it, you stop searching and compute the LCM.

Why brute-forcing is slow and error-prone

Without the LCM recognition, students often try small numbers one at a time: "Is 4 divisible by 6 and 9? No. Is 6? No. Is 8? No. Is 12? Divisible by 6 but not 9. Is 18? Divisible by 6 and 9 but not 4. Is 24? Divisible by 4 and 6 but not 9. Is 36…" This eventually gets there, but it is slow, and it is easy to forget to check the full list at each step.

With the LCM instinct:

4 = 2^2, \quad 6 = 2 \cdot 3, \quad 9 = 3^2. \qquad \operatorname{lcm} = 2^2 \cdot 3^2 = 36.

Three seconds.

Brute force versus LCM instinct on the same problemA slider scans four steps. Step 0: the problem asks for the smallest n divisible by 4, 6, and 9. Step 1: brute force tries n = 4, 6, 8, 12, 18, 24, 36 — the first n that works. Step 2: LCM instinct factors 4 = 2 squared, 6 = 2 times 3, 9 = 3 squared and reads off LCM = 2 squared times 3 squared equals 36. Step 3: side-by-side comparison shows the LCM method is about ten times faster.
The same question, two approaches. Brute force eventually lands on $36$ after seven or eight failed attempts. Factoring into primes and reading off the LCM gets there in one step and gives you the answer for *any* size of input list.

The computation itself

Once you've identified the LCM, the computation is mechanical.

Method 1 (prime factorisation). Factor each a_i into primes. For each prime p, take the maximum exponent across all the a_i's. Multiply.

\operatorname{lcm}(12, 15, 18) = \operatorname{lcm}(2^2 \cdot 3, \; 3 \cdot 5, \; 2 \cdot 3^2) = 2^2 \cdot 3^2 \cdot 5 = 180.

Method 2 (GCD identity, for two numbers). Use \operatorname{lcm}(a, b) = \frac{ab}{\gcd(a, b)}. For three or more numbers, apply iteratively: \operatorname{lcm}(a, b, c) = \operatorname{lcm}(\operatorname{lcm}(a, b), c).

\operatorname{lcm}(12, 15) = \tfrac{180}{\gcd(12, 15)} = \tfrac{180}{3} = 60; \quad \operatorname{lcm}(60, 18) = \tfrac{1080}{\gcd(60, 18)} = \tfrac{1080}{6} = 180.

Same answer. Method 1 is almost always faster for school-size numbers because factorisations are quick.

A useful variant: "smallest n leaving the same remainder"

A related pattern: find the smallest positive integer n such that n leaves a remainder of 3 when divided by 6, 9, 12.

Here, n - 3 is divisible by 6, 9, 12. So n - 3 = \operatorname{lcm}(6, 9, 12) = 36, giving n = 39. The LCM still does the heavy lifting — it is just shifted by the remainder.

What to say to yourself when you see the cue

When you read a problem, look for any of these phrasings:

If any of those patterns match, immediately stop reading for other strategies — the answer is the LCM of the relevant numbers. Spend your problem-solving budget on the factorisations, not on trying values.

The mirror-image recognition is equally useful: "largest number that divides all of these" → GCD (minimum exponent at each prime). If you see that phrasing, it is a GCD problem. Both GCD and LCM pattern-recognitions together cover a huge fraction of Olympiad and JEE-style number theory questions.

One-line takeaway

When you see "smallest n divisible by these numbers" — or any of its disguises — the answer is the LCM. Factor each number, take the maximum exponent at each prime, multiply. You are done in one line; brute force takes a page.

Related: Number Theory Basics · Factor Into Primes First · Does gcd × lcm = product Work for Three Numbers? · Euclidean Algorithm Rectangle Tiling · Coprime vs Prime