LCM and HCF
HCF is the largest number that divides all the given numbers; LCM is the smallest number divisible by all of them. HCF × LCM = product of two numbers.
Methods and formulas
| Item | How to find it |
|---|---|
| HCF by prime factors | Multiply the common primes, each at its lowest power |
| LCM by prime factors | Multiply every prime, each at its highest power |
| HCF by division | Divide the larger by the smaller, then the divisor by the remainder, until the remainder is 0; the last divisor is the HCF |
| Product rule (two numbers) | HCF × LCM = product of the two numbers |
| HCF of fractions | HCF of numerators ÷ LCM of denominators |
| LCM of fractions | LCM of numerators ÷ HCF of denominators |
| Greatest number leaving the same remainder | HCF of the differences of the numbers |
| Least number leaving remainder r with each divisor | LCM of the divisors + r |
Which one does the question want?
HCF is never larger than the smallest number; LCM is never smaller than the largest number. Use this to reject wrong options quickly.
Worked example 1: HCF and LCM of two numbers
Find the HCF and LCM of 72 and 120.
- Standard method (prime factors): 72 = 2³ × 3² and 120 = 2³ × 3 × 5.
- HCF = common primes at lowest powers = 2³ × 3 = 24. LCM = all primes at highest powers = 2³ × 3² × 5 = 360.
- Shortcut (division, then product rule): 120 ÷ 72 = 1 remainder 48; 72 ÷ 48 = 1 remainder 24; 48 ÷ 24 = 2 remainder 0, so HCF = 24. LCM = 72 × 120 ÷ 24 = 8,640 ÷ 24 = 360.
- Check: 360 ÷ 72 = 5 and 360 ÷ 120 = 3; and 24 × 360 = 8,640 = 72 × 120.
HCF = 24, LCM = 360.
Worked example 2: least number with a remainder
Find the least number which, when divided by 12, 15 and 20, leaves a remainder of 5 in each case.
- Standard method: 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5, so LCM = 2² × 3 × 5 = 60. The number must be 60 + 5 = 65.
- Shortcut (multiples of the largest): test multiples of 20 (20, 40, 60) until one is divisible by 12 and 15: 60. Add the remainder: 65.
- Check: 65 = 12 × 5 + 5 = 15 × 4 + 5 = 20 × 3 + 5.
65.
Worked example 3: equal pieces (HCF word problem)
Three ropes are 84 m, 126 m and 210 m long. They must be cut into equal pieces, as long as possible, with nothing left over. Find the length of each piece and the number of pieces.
- Standard method: 84 = 2² × 3 × 7, 126 = 2 × 3² × 7, 210 = 2 × 3 × 5 × 7. HCF = 2 × 3 × 7 = 42 m.
- Pieces = 84 ÷ 42 + 126 ÷ 42 + 210 ÷ 42 = 2 + 3 + 5 = 10.
- Shortcut (division): 126 ÷ 84 = 1 remainder 42; 84 ÷ 42 = 2 remainder 0, so HCF(84, 126) = 42. 210 ÷ 42 = 5 exactly, so the HCF stays 42. Pieces = total length ÷ 42 = 420 ÷ 42 = 10.
- Check: 10 pieces × 42 m = 420 m = 84 + 126 + 210.
Each piece is 42 m; there are 10 pieces.
Bank, SSC and railway papers ask bells ringing together, traffic lights, tiles and ropes as word problems, plus the HCF × LCM = product check. Placement tests mix in fractions, such as the HCF of 2/3, 4/9 and 8/15.
2 marks: “The HCF of two numbers is 12 and their LCM is 360. If one number is 72, find the other.” Other number = HCF × LCM ÷ given number = 12 × 360 ÷ 72 = 60.
5 marks: “Find the least number which, when divided by 6, 8 and 10, leaves a remainder of 3 in each case.” LCM(6, 8, 10) = 120; required number = 120 + 3 = 123. Check: 123 = 6 × 20 + 3 = 8 × 15 + 3 = 10 × 12 + 3.
10 marks: “Explain the methods of finding HCF and LCM, and solve one word problem on each.” Prime-factor and division methods with 72 and 120; product rule; fractions; then a rope (HCF) problem and a bells or remainder (LCM) problem with checks.
Greatest → HCF; least → LCM. For two numbers, HCF × LCM = product. Same remainder → HCF of differences; given remainder → LCM + r.