LCM & HCF

1HCF and LCM, read from the primes

The HCF (also GCD) of a set of numbers is the largest positive integer that divides every one of them; the LCM is the smallest positive integer that every one of them divides. Both are read off a complete prime factorisation: for each prime, the HCF keeps the lowest power that appears and the LCM keeps the highest.

So for 12 = 2² × 3 and 18 = 2 × 3² the shared floor is 2¹ × 3¹ = 6 and the shared ceiling is 2² × 3² = 36. Listing factors and multiples by hand reaches the same two answers, and takes longer the moment a third number arrives.

Figure. Lengths are to scale for 12 and 18. The HCF bar is the longest integer length that tiles both 12 and 18 with no remainder; the LCM bar is the shortest integer length that both 12 and 18 tile with no remainder. The same two answers come from the min/max exponents in the ledger.

How it works

1. Factorise each numberStrip primes until every factor is prime; write the result as a product of prime powers.
2. Take floors and ceilingsPer prime, the HCF uses the smallest exponent present and the LCM the largest. A prime missing from a number contributes exponent 0 to the HCF.
3. Multiply backThe product of those chosen powers is the HCF or the LCM. Check on a small pair before trusting a three-number answer.

2HCF × LCM equals the product — for a pair

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b. Rearranged, any one of the four quantities is the product of the other three divided by the remaining known value, which is how most "find the other number" questions are finished in one line.

The identity is false for three or more numbers: HCF(6, 10, 15) = 1 and LCM(6, 10, 15) = 30, but 6 × 10 × 15 = 900, not 30. When a question hands you three numbers, factorise; do not reach for the product.

Figure. Both panels have the same area 216: 12\times 18 on the left and \mathrm{HCF}\times\mathrm{LCM}=6\times 36 on the right. That is why any one of the four quantities is the product of the other three divided by the remaining known value — and why the identity stops at pairs.

How it works

1. Confirm there are two numbersThe product relation applies to a pair. Three given values that are not "HCF, LCM, one number" need prime factors instead.
2. Write the productHCF × LCM = a × b, so the unknown is (product of the three knowns) ÷ (the remaining known).
3. Sanity-checkThe HCF must divide both numbers, and both numbers must divide the LCM. If either fails, an input was misread.

3HCF and LCM of fractions

Fractions swap the roles of the denominator. The HCF of a set of fractions is (HCF of the numerators) / (LCM of the denominators), and the LCM of the fractions is (LCM of the numerators) / (HCF of the denominators). Both formulas assume the fractions are in lowest terms; cancel each one before applying them.

The two formulas are mirrors, not twins: whichever operation you apply to the numerators, you apply the other operation to the denominators. Mixing the two — LCM over LCM, or HCF over HCF — is the standard wrong answer.

Figure. Fraction HCF puts HCF on the numerators and LCM on the denominators; fraction LCM does the reverse. The denominators swap roles — that is the whole rule, once every fraction is in lowest terms.

4Largest divisor that leaves the same remainder

If a number d divides a, b and c leaving the same remainder r in each case, then d divides each of a − r, b − r and c − r, so the largest such d is HCF(a − r, b − r, c − r). The remainder must be smaller than that answer — if the HCF comes out equal to r, the numbers were exact multiples and the remainder was really 0.

When the remainder is the same but unknown, subtract the numbers from each other instead: the largest d is the HCF of the pairwise differences. Those differences cancel r, so you never need its value.

Figure. Each bar keeps the same remainder strip r=2. Strip it off and the bodies 60, 130, 235 share divisor d=\mathrm{HCF}=5 — the largest number that leaves remainder 2 with each original.

How it works

1. Read whether r is known"Leaving remainder 2" names r. "Leaving the same remainder in each case" does not — use differences.
2. Subtract, then take HCFKnown r: HCF(a − r, b − r, …). Unknown r: HCF of the pairwise differences of the given numbers.
3. Check the remainderDivide one of the originals by the answer and confirm the remainder is what the question claimed (or confirm it matches across the set).

5Smallest number that leaves a fixed remainder

The smallest positive number that leaves remainder r when divided by each of a, b, c, … is LCM(a, b, c, …) + r — provided r is smaller than every divisor. Every common multiple of the divisors leaves remainder 0; shifting that multiple up by r leaves remainder r with each of them.

This is the LCM twin of the previous concept. Same-remainder questions that ask for a largest divisor are HCF questions; same-remainder questions that ask for a smallest number meeting a list of divisions are LCM questions. The wording "largest" versus "smallest" is the whole tell.

Figure. Every common multiple of 8, 12 and 18 is a multiple of their LCM 72. Shift that multiple up by the fixed remainder 7 to get N=79 — the smallest positive number that leaves remainder 7 with each divisor.

How it works

1. Confirm the shapeThe question wants a number N such that N ÷ a, N ÷ b, … all leave the same remainder r.
2. Take the LCMLCM of the divisors is the smallest N that leaves remainder 0 with each.
3. Add the remainderN = LCM + r. Check that r is smaller than every divisor, or the division algorithm would have carried.

6When periodic events next coincide

If events repeat every a, b, c, … units and they occur together at a start time, they next occur together after LCM(a, b, c, …) units. Bells, flashing lights, running tracks and "two people meeting again at the starting point" are the same question in different clothes.

Take the highest power of each prime across the intervals — that is the LCM — and add it to the start time if the question asks for a clock reading rather than a duration. Intervals given in mixed units must be converted to one unit before the LCM is formed.

Figure. Each bar is one bell's interval, drawn to a common scale against the 72 s return. 72 is an integer multiple of every interval (12×6, 9×8, 6×12, 4×18) and no smaller positive time is, which is exactly what the LCM asserts.

How it works

1. Name the intervalsWrite every period in the same unit. Discard any "they start together" colour that is not a number.
2. Form the LCMHighest power of each prime across the intervals. That duration is the first return.
3. Add the start if askedA clock-time question wants start + LCM, reduced mod 12 or 24 as the options demand.