HCF & LCM
🔒 Log in to trackHighest common factor and lowest common multiple — a small but near-guaranteed slot in CGL. Most questions are template word problems: greatest divisor with remainders (HCF of differences), least number with given remainders (LCM + r), bells, tiles and ratio pairs.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
56 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (56 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Greatest number dividing with same/known remainders (HCF of differences)
very common'Find the greatest number which divides x, y, z leaving remainders a, b, c' (remainders equal or different).
How to solve: Subtract each remainder from its number and take the HCF of the results — the divisor divides every (number − remainder) exactly. If the remainders are not even mentioned, take the HCF of the pairwise differences of the numbers instead.
Example: Find the greatest number which divides 82, 150 and 233 leaving remainders 4, 7 and 12 respectively.
Subtract: 78, 143, 221. HCF = 13 (each is a multiple of 13).
Least number leaving the same remainder (LCM + r)
very common'Find the least number which when divided by a, b, c leaves remainder r in each case.'
How to solve: Take the LCM of the divisors and add r. With an extra condition (n-digit bound, or divisible by p), write N = LCM·k + r and fit the condition by testing k = 1, 2, 3, …
Example: Find the least number which when divided by 8, 12, 15 and 20 leaves remainder 4 in each case.
LCM(8, 12, 15, 20) = 120, so the answer is 120 + 4 = 124.
Remainder is (divisor − c) each time (LCM − c)
commonThe remainders look different, but each is a fixed amount less than its divisor — e.g. remainders 2, 3, 4 with divisors 3, 4, 5.
How to solve: Check that divisor − remainder is the same number c in every case. Then N + c is divisible by all divisors, so the least N is LCM − c (and other values are LCM·k − c).
Example: Find the least number which when divided by 6, 9 and 12 leaves remainders 5, 8 and 11 respectively.
Each remainder is 1 less than its divisor: N = LCM − 1 = 36 − 1 = 35.
Ratio + HCF/LCM → find the numbers
very common'Two numbers are in ratio m : n and their HCF (or LCM) is …' — then sum, difference, product or LCM is asked.
How to solve: Cancel the ratio to co-prime parts m, n. Numbers are h·m and h·n where h is the HCF. From the LCM: h = LCM/(m·n). Then compute whatever is asked — sum = h(m + n), difference = h(n − m), LCM = h·m·n.
Example: Two numbers are in the ratio 5 : 7 and their LCM is 140. Find the numbers.
h = 140/(5 × 7) = 4 → numbers 20 and 28.
Product / HCF / LCM / one-number relation
commonTwo of {product, HCF, LCM, one number} are given; find the fourth, or 'how many pairs are possible'.
How to solve: Use HCF × LCM = product (two numbers only). For pair counts, split product ÷ HCF² (or LCM ÷ HCF) into co-prime factor pairs, always including the pair (1, M).
Example: The product of two numbers is 2160 and their HCF is 12. Find their LCM.
LCM = 2160/12 = 180.
Bells / lights / runners together again
commonEvents repeating at fixed intervals start together; asked when they next coincide, or how many times inside a time window.
How to solve: Convert all intervals to one unit, take the LCM — that is the coincidence gap. Add it to the start time for 'when next'; divide the window by it for 'how many times' (add 1 only if the start itself counts).
Example: Three bells toll at intervals of 9, 12 and 15 minutes, together at 8 a.m. When do they next toll together?
LCM = 180 minutes = 3 hours → 11 a.m.
Largest tile / rod / measure / group size (HCF)
commonPaving with the largest square tiles, cutting equal planks of greatest length, the biggest vessel filling containers, biggest equal groups.
How to solve: Convert all measurements to the same unit and take the HCF. Then count: tiles = (L × B)/side², pieces = total length ÷ HCF, groups = count ÷ HCF.
Example: What is the largest vessel that can fill 144 L, 180 L and 240 L containers exactly?
HCF(144, 180, 240) = 12 litres.
Greatest / least n-digit number with LCM or remainder conditions
common'Find the greatest four-digit number divisible by …' / 'least five-digit number which divided by … leaves remainder …'.
How to solve: Take the LCM first. Least: step up from 10…0 to the next multiple (fit N = LCM·k + r into the range if a remainder is given). Greatest: reduce 99…9 by its remainder.
Example: Find the least five-digit number exactly divisible by 32, 36, 40, 45 and 48.
LCM = 1440; 10000 ÷ 1440 leaves 1360 → add 80 → 10080.
HCF / LCM of fractions or decimals
occasionalA list like 2/3, 8/9, 10/27, or decimals like 0.54, 1.8, 7.2.
How to solve: Fractions: HCF = (HCF of numerators)/(LCM of denominators); LCM is the mirror image. Decimals: multiply everything by the same power of 10 to clear the point, solve as integers, then put the point back.
Example: Find the HCF of 0.54, 1.8 and 7.2.
As 54, 180, 720 the HCF is 18 → answer 0.18.