LCM

1. Meaning and relationship with HCF

A multiple of a positive integer is obtained by multiplying it by an integer. The positive multiples of 6 are 6, 12, 18, 24 and so on; those of 8 are 8, 16, 24, 32 and so on. Their first common positive multiple is 24, so LCM(6, 8) = 24. Every other common positive multiple is a multiple of 24. Although zero is divisible by every positive integer, it is excluded when identifying the least positive common multiple.

The distinction between LCM and highest common factor, also called greatest common divisor, is fundamental. HCF identifies the largest number that divides all the given numbers exactly. LCM identifies the smallest positive number that is divisible by all of them. Thus, the largest equal length used to cut several ropes usually involves HCF, whereas the smallest total that can be arranged into several different group sizes usually involves LCM.

For two positive integers, their product equals the product of their HCF and LCM. For example, 18 × 24 = 6 × 72. This identity is not generally valid for three or more numbers. Also, the LCM is at least as large as the largest given number. If the largest number is divisible by every other number, it is already the LCM.

2. Reliable methods of calculation

Listing multiples is useful when the numbers are small and the common multiple appears quickly. For 4 and 10, compare 4, 8, 12, 16, 20 with 10, 20; the answer is 20. However, listing becomes inefficient for larger numbers or several inputs. In such cases, prime factorisation or the division method is more systematic and reduces missed factors.

Under prime factorisation, express each number as a product of primes and select the highest exponent of every prime present. For 72 = 2³ × 3², 120 = 2³ × 3 × 5 and 150 = 2 × 3 × 5², the LCM is 2³ × 3² × 5² = 1,800. In contrast, the HCF uses only primes shared by every number, with their lowest exponents.

In the division or ladder method, divide by a prime that divides at least one entry, carrying down entries not divisible by it. Continue until every entry becomes 1, then multiply all the divisors used. Another efficient route is LCM(a, b) = (a ÷ HCF(a, b)) × b. For several numbers, proceed pairwise: LCM(a, b, c) = LCM(LCM(a, b), c).

• If the numbers are pairwise coprime, their LCM is their product.
• An overall HCF of 1 does not imply pairwise coprimality: 6, 10 and 15 have HCF 1 but LCM 30, not 900.
• For a positive integer k, LCM(ka, kb) = k × LCM(a, b).

3. Recurring events and time-based problems

LCM is especially useful when periodic events begin together and repeat at fixed intervals. If three bells ring every 12, 18 and 30 minutes and ring together at 9:00 a.m., their next joint ringing occurs after LCM(12, 18, 30) = 180 minutes, at noon. The answer is an elapsed duration first; convert it into a clock time only after the LCM has been calculated.

Always express intervals in the same unit before calculating. For events occurring every 45 seconds and 1 minute, use 45 and 60 seconds; their LCM is 180 seconds, or 3 minutes. Counting coincidences requires attention to endpoints. If the joint interval is L and observation begins with a coincidence, the number in the inclusive interval from time 0 to time T is floor(T ÷ L) + 1.

Different starting times change the problem. An event occurring at times 0, 6, 12 and so on and another occurring at times 2, 10, 18 and so on first coincide at time 18, not at LCM(6, 8) = 24. Similarly, runners completing laps in 60 and 90 seconds return together to the starting point after 180 seconds, but their first meeting anywhere must be checked using direction and relative speed.

4. Divisibility, remainders and fractions

When a number N leaves the same remainder r on division by several divisors, N − r is divisible by all those divisors. Therefore, N = kL + r, where L is their LCM. If division by 5, 7 and 9 leaves remainder 2, then N = 315k + 2. However, if the question merely asks for the least positive number, 2 itself qualifies. The answer becomes 317 only when a condition such as being greater than every divisor excludes 2.

If a number falls short of a multiple by the same amount for every divisor, add that amount. For example, remainders 4, 6 and 8 on division by 5, 7 and 9 respectively mean that N + 1 is divisible by all three. Hence N = 315k − 1, giving the least positive solution 314. Arbitrary unequal remainders cannot be handled by automatically adding or subtracting one common quantity.

LCM also supplies the least common denominator for adding fractions. The denominators 12 and 18 have LCM 36, so 5/12 + 7/18 = 15/36 + 14/36 = 29/36. If a question defines a common multiple of fractions as a quantity that is an integer multiple of each, first reduce the fractions: their LCM equals the LCM of numerators divided by the HCF of denominators.

5. CSAT problem-solving strategy and checks

Start by identifying the required quantity: an equal grouping size, a common total, an elapsed interval or an unknown number satisfying remainders. Words such as smallest, together and exactly divisible can suggest LCM, but they are not sufficient by themselves. Translate the statement into a divisibility condition before calculating. This prevents confusion between LCM, HCF and relative-speed problems.

Remove redundant inputs where possible. In LCM(8, 16, 24), the 8 adds no restriction because every multiple of 16 is already divisible by 8; calculate LCM(16, 24) = 48. For the smallest four-digit number divisible by 12, 18 and 30, first find L = 180. The first multiple of 180 at least 1,000 is 180 × 6 = 1,080.

Finally, verify the answer against every original condition. Check divisibility, permitted remainders, units and inclusion of the initial event. In multiple-choice questions, testing options can be faster than full factorisation, particularly when options are few and ordered. Nevertheless, proving that an option is divisible by all inputs establishes only that it is a common multiple; minimality must also be checked.

Previous year questions

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Practice questions