HCF (Highest Common Factor) and LCM (Lowest Common Multiple) are two of the most reliable marks-sources in SSC CGL Quant — not because the concepts are hard, but because there are exactly four or five question types that repeat across years, and once you know them, each question takes under 60 seconds.
Here is what the two terms actually mean, stripped of textbook language.
HCF is the largest number that divides two or more given numbers without leaving a remainder. Think of it as the biggest "common ruler" you can use to measure two lengths exactly. If you have ropes of 36 cm and 60 cm and you want to cut them into equal pieces with no waste, the longest possible piece is HCF(36, 60) = 12 cm.
LCM is the smallest number that is divisible by two or more given numbers. Think of it as the earliest "common meeting point." If one event repeats every 4 days and another repeats every 6 days, the first day they coincide again is LCM(4, 6) = 12 days from now.
The two concepts are not independent — they are connected by one single master property:
This holds for exactly two numbers. For three or more, this formula does not apply directly — a mistake that appears repeatedly in CGL answer sheets.
An analogy that makes this click: HCF is the intersection of prime-factor sets (take the minimum power of each common prime), and LCM is the union (take the maximum power of every prime that appears). Once you see it this way, you will never confuse which operation gives which result.
Break each number into prime factors. Then:
Example — find HCF and LCM of 36, 60, and 84.
HCF: 2 and 3 appear in all three. Lowest powers: 2² and 3¹.
LCM: All primes present: 2, 3, 5, 7. Highest powers: 2², 3², 5¹, 7¹.
For large numbers where factorization is slow, use repeated division (also called Euclid's algorithm).
To find HCF(168, 252):
252 ÷ 168 = 1 remainder 84168 ÷ 84 = 2 remainder 0This is typically faster than full factorization for numbers above 100.
This is the most-tested formula in SSC CGL for this topic. Know it cold.
Rearrangements you will use:
HCF × LCM = a × bWhen two numbers are in ratio m : n, write them as mk and nk. Their HCF is always k (assuming m and n are coprime, i.e., share no common factor).
The LCM of mk and nk = mnk (when gcd(m,n) = 1).
So: LCM = m × n × k and HCF = k.
Given LCM, solve for k, then find the numbers.
HCF must always be a factor of LCM. This is non-negotiable. If you're given a pair (H, L) and H does not divide L exactly, that pair is invalid — no two numbers can have those as their HCF and LCM. This generates one of CGL's favorite "which cannot be" question types.
When events happen at intervals of a, b, c seconds/minutes, and you want the next time they all coincide: find LCM(a, b, c) and add it to the starting time. This is the bell-problem template that has appeared multiple times in CGL.
A typical phrasing: "Find the greatest number that when subtracted from N gives a result divisible by each of p, q, r."
Step 1: Find LCM(p, q, r). Call it L.
Step 2: Find the largest multiple of L that is ≤ N. Call it M.
Step 3: Answer = N − M.
Why? Because N − answer must be divisible by L. The answer is smallest when N − answer is the largest multiple of L not exceeding N.
Any time a question gives you HCF, LCM, and one number — apply x = (HCF × LCM) ÷ known number directly. No factorization needed.
Example: HCF = 12, LCM = 180, one number = 36. Other number = (12 × 180) ÷ 36 = 2160 ÷ 36 = 60.
Standard method (factorize both, compare): ~45 seconds. This formula: ~12 seconds. The only trap is applying this to three numbers — it doesn't work there.
In "which cannot be HCF and LCM" questions, test each option by dividing LCM ÷ HCF. If it's not a whole number, that pair is invalid — stop there, mark it, move on.
Example: Option (35, 120) → 120 ÷ 35 = 3.43 (not whole) → invalid. Done in 5 seconds vs. trying to construct actual numbers (~40 seconds).
Numbers in ratio m:n → write as mk and nk. If gcd(m,n) = 1, then LCM = mnk.
Given ratio 3:4 and LCM = 84: m=3, n=4, mn=12. So 12k = 84, k=7. Numbers = 21 and 28.
Standard method (guess-and-check multiples): ~35 seconds. This approach: ~10 seconds. Works cleanly when m and n are coprime — check that first.
Find LCM of intervals in seconds. Convert to minutes + seconds only after computing.
LCM(24, 36, 54): Factor → 24 = 2³×3, 36 = 2²×3², 54 = 2×3³. LCM = 2³×3³ = 8×27 = 216 seconds. Convert: 216 = 3 min 36 sec. Add to start time.
The mistake most test-takers make is converting to minutes before finding LCM, then getting confused with mixed units. Keep everything in seconds until after the LCM calculation. Saves one unit-conversion error and ~15 seconds of re-checking.
Template: subtract from N so the result is divisible by each of p, q, r.
Result must be a multiple of LCM(p,q,r). The largest such multiple ≤ N is floor(N ÷ LCM) × LCM. Subtract that from N.
For N=5834, LCM(20,28,32,35)=1120: floor(5834÷1120)=5, nearest multiple=5600. Answer = 5834−5600 = 234. But check: 5834−4714=1120, which IS divisible. So the question asks for the "greatest number subtracted," meaning the answer is the difference 5834−1120=4714. Re-read: you subtract 4714 to get 1120, which is divisible. Standard mis-read costs 30 seconds; this template keeps you on track.
In the exam hall, classify the question in the first five seconds:
Given HCF + LCM + one number → find the other: Use x = (HCF × LCM) ÷ given number. Done.
"Which pair cannot be HCF and LCM?": Check each option — does HCF divide LCM? If no, that's the answer. Start with the option that looks most suspicious (non-round numbers).
Two numbers in ratio m:n, LCM given: Write numbers as mk, nk. LCM = mnk (if gcd(m,n)=1). Solve for k.
Bell / simultaneous events: Find LCM of all intervals (in the same unit — seconds preferred). Convert to readable time. Add to start.
Greatest/smallest number subtracted or added: Find LCM of divisors. Locate the nearest multiple of LCM above or below the given number. The difference is your answer.
If none of the above fits, fall back to prime factorization. Do not use long division for numbers under 100 — factorization is faster there.
Why this question: The classic validity-check question. Every CGL batch since 2010 has seen at least one variant of this.
Solving path: For each option, divide LCM by HCF. Option A: 96÷12=8 (whole, valid). Option B: 120÷35=3.43 (not whole, invalid — stop here). Option C: 72÷18=4 (valid). Option D: 90÷15=6 (valid). Answer is B. Time: under 20 seconds if you go in order.
Why this question: Bell problems are a CGL staple. The trap is unit conversion — candidates who convert to minutes before computing LCM often get a wrong LCM.
Solving path: Work in seconds throughout. 24 = 2³×3, 36 = 2²×3², 54 = 2×3³. LCM = 2³×3³ = 216 seconds. 216 ÷ 60 = 3 minutes 36 seconds. 10:15:00 + 3 min 36 sec = 10:18:36 a.m.
Why this question: The HCF×LCM property question, first-principles version. Straightforward if you know the formula, a time sink if you try to find the second number by trial.
Solving path: HCF × LCM = a × b → 8 × 48 = 24 × x → 384 = 24x → x = 16. Verify: HCF(24,16)=8, LCM(24,16)=48. Correct.
Why this question: The ratio-to-LCM template. Many students guess-and-check multiples of 3 and 4, wasting 30+ seconds.
Solving path: Numbers = 3k and 4k. Since gcd(3,4)=1, LCM = 3×4×k = 12k = 84 → k=7. Numbers: 21 and 28. Greater number = 28.
Why this question: Tests the master property with verification, making it a two-step problem that weeds out candidates who stop at finding x without cross-checking.
Solving path: HCF × LCM = product → 12 × 180 = 36 × x → 2160 = 36x → x = 60. Verify: HCF(36,60): 36=2²×3², 60=2²×3×5. HCF=2²×3=12. LCM=2²×3²×5=180. Both check out.
Applying HCF×LCM = a×b to three numbers. This formula works only for two numbers. For three numbers (a, b, c), there is no direct single formula — you must use prime factorization or compute pairwise. Using a×b×c = HCF×LCM for three numbers is wrong and will give a wrong answer every time.
Confusing which gets "minimum" and which gets "maximum" power in prime factorization. HCF takes the minimum power of each common prime; LCM takes the maximum power of every prime. Swapping these gives an LCM smaller than HCF, which is a red flag you should catch before marking.
Converting units before computing LCM in bell problems. If intervals are in seconds, find LCM in seconds. Convert to minutes and seconds only after. Mixing units mid-calculation causes arithmetic errors.
Not verifying that HCF divides LCM in constructed answers. After finding numbers x and y using the master property formula, spend 5 seconds verifying that HCF(x,y) and LCM(x,y) match the given values. Especially important when options are close together (e.g., 24, 16 vs. 36, 12).
In ratio problems, forgetting to check that m and n (ratio terms) are coprime. The formula LCM(mk, nk) = mnk holds only when gcd(m,n)=1. If ratio is 4:6 (gcd=2), reduce to 2:3 first, then apply. Using the raw ratio directly gives the wrong LCM.
"Greatest number subtracted" vs. "greatest number that divides the remainder." These are two different problem types. Subtraction problems ask you to find the largest value to remove from N so the result is divisible by L. Don't confuse with "find HCF of differences" problems (which apply when you want the same remainder from multiple numbers).