LCM & HCF — RRB NTPC Mathematics
This topic carries roughly 5% of Mathematics's 30 questions. The computation (prime factorisation or the division method) is routine; the actual test is reading a word problem correctly enough to know whether it wants the LCM or the HCF.
1. What RRB NTPC actually asks
Expect direct LCM/HCF computation of two or three numbers, and word problems: "least number divisible by X, Y, Z leaving remainder r" (LCM-based), "greatest number that divides X and Y leaving a given remainder" (HCF-based), and problems using the LCM-HCF-product relationship for two numbers.
2. HCF (Highest Common Factor)
The HCF of two or more numbers is the largest number that divides all of them exactly. Find it via prime factorisation (multiply the common prime factors at their lowest shared power) or the division method (repeated Euclidean division).
3. LCM (Lowest Common Multiple)
The LCM of two or more numbers is the smallest number divisible by all of them exactly. Find it via prime factorisation (multiply every prime factor at its highest power appearing in any number).
For exactly two numbers, LCM × HCF = product of the two numbers — a fast way to find one when the other and the product are known.
4. The two classic word-problem patterns
"Least number divisible by A, B, C leaving remainder r in each case" = LCM(A, B, C) + r.
"Greatest number that divides A and B leaving remainder r in each case" = HCF(A − r, B − r).
Recognising which pattern a word problem matches is the entire skill here — once matched, the computation is routine.
Worked examples
Q1. What is the LCM of 36 and 60?
Pick an option to check your answer.
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Solution. 36 = 2² × 3², and 60 = 2² × 3 × 5. LCM takes each prime at its highest power: 2² × 3² × 5 = 4 × 9 × 5 = 180.
Verify using the product rule: HCF(36, 60) = 12, and LCM × HCF should equal 36 × 60 = 2160. Check: 180 × 12 = 2160. Answer: (b).
Q2. Find the least number which, when divided by 12, 15, and 20, leaves a remainder of 5 in each case.
Pick an option to check your answer.
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Solution. This matches the "least number divisible by A, B, C leaving remainder r" pattern: answer = LCM(12, 15, 20) + 5.
LCM(12, 15, 20) = 60 (12 = 2²×3, 15 = 3×5, 20 = 2²×5; taking each prime at its highest power gives 2²×3×5 = 60). Adding the remainder: 60 + 5 = 65. Answer: (b).
6. Common traps
- Confusing which pattern a word problem needs. "Leaves a remainder" combined with "least number divisible by" needs LCM + r; "greatest number that divides... leaving remainder" needs HCF of the reduced numbers.
- Forgetting to subtract the remainder before taking HCF in the second pattern — the HCF must be taken of (A − r) and (B − r), not of A and B directly.
- Using the product-of-numbers shortcut (LCM × HCF = product) for three or more numbers. This relationship holds only for exactly two numbers.
- Missing a prime factor's highest power when computing LCM via factorisation — double-check every prime appears in the final product at its highest power across all the given numbers.
7. Guessing strategy
If a computation is taking too long under exam pressure, a quick divisibility check on each option (does it divide the given remainder problem correctly?) can eliminate wrong options faster than fully solving from scratch.
Summary
- LCM/HCF is roughly 5% of Mathematics's 30 CBT 1 questions.
- HCF is the largest number dividing all given numbers exactly; LCM is the smallest number divisible by all of them.
- For exactly two numbers, LCM × HCF = the product of the two numbers.
- "Least number divisible by A, B, C leaving remainder r" = LCM(A, B, C) + r.
- "Greatest number dividing A and B leaving remainder r" = HCF(A − r, B − r).
- The product-rule shortcut only applies to exactly two numbers, not three or more.
