LCM
The Lowest Common Multiple (LCM) is the smallest positive number that every number in a given set divides into exactly. It looks like a school topic, but in CAT it quietly powers a whole family of problems — bells or lights that blink together, the smallest number that leaves the same remainder under several divisions, gear and circular-track meeting points, and the "find the next common event" set-ups that also surface in XAT and SNAP. The companion idea, HCF (the largest common divisor), pairs with LCM through one clean identity: for any two numbers, HCF × LCM = the product of the numbers. Master that relationship and you can recover a missing number from the other three quantities in seconds. This chapter builds LCM from prime factorisation, then layers on the HCF–LCM link, LCM of fractions, and the high-frequency CAT applications — bells ringing together and "least number leaving remainder r" problems of the form LCM·k + r. Throughout, the emphasis is on the fast factor-based method and the traps that cost careless candidates marks: confusing HCF with LCM, mishandling fractions, and forgetting that the identity HCF × LCM = product holds for exactly two numbers, not three or more.
Topics
⚡ CAT shortcuts & speed methods
The fastest ways to crack this chapter under time pressure — the techniques that separate a 95+ percentiler from the rest.
- Factorise, then take the HIGHEST power of every prime for LCM and the LOWEST power of common primes for HCF — never confuse the two.
- For two numbers only: LCM = (a×b)/HCF and the missing number = (HCF×LCM)/known. The identity HCF×LCM = a×b breaks for three or more numbers.
- HCF always divides LCM. If LCM/HCF is not an integer, the given pair is impossible — a quick sanity filter in tricky questions.
- LCM of fractions = LCM(numerators)/HCF(denominators); HCF of fractions = HCF(numerators)/LCM(denominators). Reduce each fraction first.
- Bells/lights starting together meet again after LCM(intervals); times in T seconds = T/LCM (add 1 if the starting instant is counted).
- Least number leaving the same remainder r on division by several divisors = LCM(divisors)·k + r; if every divisor is short by a constant d, the number is LCM − d.
⚠️ Common mistakes & traps
CAT is designed so that careless errors here cost you marks. Internalise each trap before the exam.
- Swapping the rules — taking lowest powers for LCM or highest powers for HCF.
- Applying HCF×LCM = product of the numbers to THREE or more numbers; the identity is valid for exactly two.
- Inverting the fraction rules — using HCF of numerators over LCM of denominators for the LCM of fractions.
- In remainder problems, adding the remainder to each number instead of to the single LCM, or forgetting the LCM − d case when divisors fall short by a constant.
- Counting the starting instant when the question asks "how many times after they start", which double-counts one toll/flash.
📈 CAT exam insight & PYQ analysis
🎴 Flashcards — instant recall
Tap a card to reveal the answer. Drill these until they are automatic.
📌 Quick revision
Chapter test
🏆 Vidaara CAT success checklist
You have truly mastered LCM when you can tick every box below.
- Recall every formula in this chapter without looking them up
- Solve each topic’s practice set with at least 80% accuracy
- Use the chapter shortcuts to cut your solving time in half
- Spot and avoid every common trap listed above
- Score 80%+ on the timed chapter test
📋 Chapter mastery scorecard
Track where you stand. Aim for the target before moving to the next chapter.
| Skill checkpoint | Target |
|---|---|
| Concept theory & formulas understood | 100% |
| Topic practice sets attempted (3 topics) | 3/3 |
| Best topic-test score | — → 80%+ |
| Chapter test score | — → 80%+ |
| Flashcards drilled to instant recall | 12 cards |
Formula Reference Sheet
Core LCM & HCF
| LCM by factorisation | LCM = product of each prime to its HIGHEST power |
|---|---|
| HCF by factorisation | HCF = product of each common prime to its LOWEST power |
| HCF–LCM identity (two numbers) | HCF(a,b) × LCM(a,b) = a × b |
| Missing LCM | LCM = (a × b) / HCF |
| Missing number | b = (HCF × LCM) / a |
Fractions & applications
| LCM of fractions | LCM(numerators) / HCF(denominators) |
|---|---|
| HCF of fractions | HCF(numerators) / LCM(denominators) |
| Bells ring together | gap = LCM of the individual intervals |
| Least number, same remainder r | N = LCM(divisors) × k + r |
| Least number, exactly divisible | N = LCM(divisors) (the case r = 0) |