HCF (Highest Common Factor) and LCM (Least Common Multiple) are two of the most reliably tested topics in UP Police Constable quant. If you have taken even a single mock, you have seen at least one question from this chapter.
Here is the plain-language version of each:
HCF is the largest number that divides two or more given numbers without leaving a remainder. Think of it as the biggest "tile" that fits evenly into two rooms of different sizes — if one room is 12 metres wide and another is 18 metres wide, the biggest tile that fits both without cutting is 6 metres. That 6 is the HCF.
LCM is the smallest number that is divisible by two or more given numbers. If two buses leave a stop every 12 minutes and every 18 minutes respectively, the LCM (36 minutes) tells you when both buses will be at the stop together again.
A fact you must tattoo in your memory:
This holds for exactly two numbers. The moment you see three numbers, this formula does not apply directly — use it only pair-wise or via prime factorisation. Most UP Police questions involve two numbers, so this formula is your primary weapon.
One more concept worth naming explicitly: the "same remainder" problem. You will be asked for the largest number that divides a set of numbers leaving the same remainder. That is always the HCF of the pairwise differences of those numbers — a specific application we will cover in the Deep Dive.
Break every number into its prime factors.
Example: Find HCF and LCM of 135, 171, 192.
Common prime factor: 3, lowest power = . So HCF = 3.
For LCM: take , , , — but note the question only asks HCF here. This illustrates how quickly you can eliminate wrong answers once you spot the common factor.
This is faster than prime factorisation for large ugly numbers.
Steps:
Example: HCF of 234 and 452.
HCF = 2. This matches the prime factorisation route but is more systematic for large numbers.
For two numbers a and b:
Nearly half of UP Police HCF-LCM questions reduce to one of these two formulae. Drill them until they are automatic.
If two numbers are in ratio m : n and their HCF is h, then:
m × hn × hm × n × hThis is a direct consequence of the fact that when you strip the HCF out of both numbers, what remains must be coprime (no common factors). The LCM is then HCF × product of the ratio parts.
Example: Numbers in ratio 3:4, HCF = 88.
You do not even need to compute the actual numbers — just multiply the ratio parts with the HCF.
"Find the largest number that divides 1005, 1245, and 1485 leaving the same remainder."
Rule: Subtract numbers pairwise, then find HCF of the differences.
HCF of 240, 240, 480:
Answer: 240.
Why does this work? If a number d divides both A and B leaving the same remainder r, then d divides (A - r) and (B - r), which means d divides (A - B). The largest such d is the HCF of all differences.
You may see these, though they are less common in UP Police:
When numbers are in ratio m:n with HCF h, write LCM = m × n × h directly. No need to compute individual numbers first.
Worked example: ratio 3:4, HCF 88 → LCM = 3 × 4 × 88 = 1056.
Standard method (find both numbers, use formula): ~45 seconds. This pattern: ~10 seconds. You save one full multiplication step and a division.
Whenever the question gives you HCF, LCM, and one number and asks for the other, immediately write: Other number = (HCF × LCM) ÷ First number. Do not attempt prime factorisation — that will cost you 60–90 seconds on large numbers.
Example: HCF = 4, LCM = 5940, one number = 108. Other = (4 × 5940) ÷ 108 = 23760 ÷ 108 = 220. Three arithmetic steps, under 20 seconds.
Same remainder questions always reduce to: find HCF of pairwise differences.
If three numbers are given, check if consecutive differences are equal first. If they are (like 240, 240 in the example above), the HCF is the difference itself. This saves computing the full HCF chain. Standard method: ~50 seconds. Spotting equal differences: ~8 seconds.
For two large numbers in a PYQ (like 1002 and 1314 where HCF is given), you do not need to verify HCF at all — the question already tells you it is 6. Just use LCM = (1002 × 1314) ÷ 6. Break it as: (1002 ÷ 6) × 1314 = 167 × 1314. Then compute: 167 × 1314 = 167 × 1000 + 167 × 314 = 167000 + 52438 = 219438.
This split-and-multiply approach avoids working with the full six-digit product. Standard multiplication: 6 steps. Split approach: 4 steps, ~30 seconds faster.
HCF must divide every number in the set. Quickly check which answer options actually divide all given numbers. In the HCF(234, 452) question, 234 is even (ends in 4) and 452 is even — so HCF must be even. Options 16, 32, 8 are even but do they divide 234? 234 ÷ 8 = 29.25 — not exact. So eliminate 8, 16, 32 without full factorisation. Only 2 remains. Time: ~15 seconds versus factorisation: ~40 seconds.
Read the question and identify the type in 5 seconds:
LCM = (a × b) ÷ HCF. Do not factorise.Second = (HCF × LCM) ÷ First.LCM = m × n × h. One multiplication.If you cannot identify the type within 5 seconds, scan the answer options — if they are close to each other, use elimination via divisibility. If they are spread far apart, your calculation is probably on track.
Why this question: Tests the basic LCM formula when HCF is given — the most common question type in this chapter.
Solving path: HCF = 6 is given. Use LCM = (1002 × 1314) ÷ 6. Divide one number by 6 first: 1002 ÷ 6 = 167. Then 167 × 1314: split as 167 × 1000 + 167 × 314 = 167000 + 52438 = 219438. Match to option C.
Why this question: Ratio-based HCF-LCM — tests whether you know the LCM = m × n × h shortcut.
Solving path: Numbers in ratio 3:4, HCF = 88. Apply directly: LCM = 3 × 4 × 88 = 12 × 88 = 1056. No need to find individual numbers. Match to option D.
Why this question: "Greatest number that divides both" = HCF. Tests whether you recognise the phrasing.
Solving path: Check answer options via divisibility. Does 17 divide 187? 187 ÷ 17 = 11. Yes. Does 17 divide 391? 391 ÷ 17 = 23. Yes. No other option works. Answer: 17.
Why this question: Same-remainder problems — the trickiest phrasing in this chapter.
Solving path: Subtract consecutive numbers: 1245 - 1005 = 240, 1485 - 1245 = 240. Differences are equal, so HCF of differences = 240 directly. Answer: 240. Verify: 1005 ÷ 240 = 4 remainder 45; 1245 ÷ 240 = 5 remainder 45. Confirmed.
Why this question: Tests the "find the second number" formula with a three-digit divisor — arithmetic precision matters.
Solving path: Other number = (HCF × LCM) ÷ First = (4 × 5940) ÷ 108 = 23760 ÷ 108. Simplify: 23760 ÷ 108 = 2640 ÷ 12 = 220. Answer: 220.
Applying the product formula to three numbers. The formula HCF × LCM = a × b works only for two numbers. For three numbers, you must use prime factorisation. Writing HCF × LCM = a × b × c is incorrect and will give a wrong answer every time.
Confusing HCF and LCM in the ratio trick. When numbers are in ratio m:n with HCF h, the LCM is m × n × h, not m + n × h or just m × n. Students who recall the formula vaguely multiply the wrong things.
Missing the "same remainder" signal. The phrase "leaves the same remainder" or "समान शेषफल" means you must take differences first, then HCF. Students who skip this step and directly find HCF of the original numbers get a wrong answer.
Not simplifying before multiplying. In (4 × 5940) ÷ 108, many students compute 4 × 5940 = 23760 and then try to divide 23760 by 108 in one shot, making arithmetic errors. Instead, cancel first: 5940 ÷ 108 = 55, then 4 × 55 = 220. Always simplify before you multiply.
Treating "HCF" and "GCD" as different things. They are the same. Questions may use either term. The method and answer are identical.
Ignoring the option-elimination route for HCF questions. When you need HCF and options are given, a valid HCF must divide every number in the set. Checking divisibility of options is often faster than running the full Euclid algorithm. Train yourself to look at options before computing.