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Numerical Aptitude Syllabus / Module 2 · HCF & LCM / Week 2

Week 2: HCF, LCM & Factors

HCF and LCM are two questions about the same prime factorization, asked in opposite directions — and once that's clear, most of this chapter's "different" question types turn out to be the same calculation wearing a different word problem. This week builds fast prime factorization, both standard methods, and the identity that connects the two.

Module 2 of 20 Week 2 of 20 ~2–3 Hours Practice Problems Included

By the end of this week, you'll be able to

  • Prime-factorize a number quickly, without a factor table
  • Find HCF and LCM by factorization and by the division method
  • Recognize an HCF or LCM word problem from its phrasing alone
1 Fast Prime Factorization 2 HCF & LCM: Two Methods 3 The HCF × LCM Identity 4 Recognizing Word Problems 5 Practice Problems 6 Knowledge Check

1. Fast Prime Factorization

Every method in this chapter starts from the same first step: break a number into its prime factors. Doing this quickly, rather than testing every number one by one, comes down to trying small primes in order and using Week 1's divisibility rules to skip ahead.

factorizing 396
396 -> even -> divide by 2: 198
198 -> even -> divide by 2: 99
99  -> digit sum 18, div by 3 -> divide by 3: 33
33  -> digit sum 6, div by 3  -> divide by 3: 11
11  -> prime -> stop

396 = 2^2 x 3^2 x 11

You only need to test primes up to √n — if no prime up to the square root divides the number, it's prime. For most test-level numbers (under a few thousand), that means checking 2, 3, 5, 7, 11, 13... rarely further, which keeps this fast even without memorized factor tables.

2. HCF & LCM: Two Methods

HCF (Highest Common Factor) is the largest number that divides every given number; LCM (Least Common Multiple) is the smallest number every given number divides into. Both can be found two ways.

method 1 — by factorization
Find HCF and LCM of 36 and 60.

36 = 2^2 x 3^2
60 = 2^2 x 3 x 5

HCF = product of the SMALLEST power of each COMMON prime
    = 2^2 x 3^1 = 4 x 3 = 12

LCM = product of the LARGEST power of every prime that
      appears in EITHER number
    = 2^2 x 3^2 x 5 = 4 x 9 x 5 = 180
method 2 — division method (faster for 3+ numbers)
Find HCF of 36, 60 and 84 by repeated division:

  36 = 60 x 0 + 36  ->  gcd(36,60):
  60 = 36 x 1 + 24
  36 = 24 x 1 + 12
  24 = 12 x 2 + 0    ->  gcd(36,60) = 12

  now gcd(12, 84):
  84 = 12 x 7 + 0    ->  gcd = 12

HCF of all three = 12

Factorization is more intuitive and works well for two numbers with small factors; the division method (Euclid's algorithm) is faster once numbers get large or ugly to factorize, since it never requires finding the actual prime factors at all.

3. The HCF × LCM Identity

For exactly two numbers, HCF and LCM aren't independent — they're linked by one identity worth having memorized, since it turns a two-step calculation into one:

the identity (two numbers only)
HCF(a, b) x LCM(a, b) = a x b

Check against Section 2's example:
  HCF(36, 60) x LCM(36, 60) = 12 x 180 = 2,160
  36 x 60                    = 2,160   -> matches

This means if a question gives you the product of two numbers and one of HCF/LCM, you can find the other directly — a very common test-question shape. It does not extend to three or more numbers in the same simple form; for three numbers, HCF × LCM ≠ product in general, which is a trap worth remembering explicitly.

Two numbers with a given HCF must be multiples of it

If two numbers have HCF = h, they can always be written as h×m and h×n where m and n share no common factor (are coprime) — this parametrization is the fastest way to solve "find two numbers whose HCF is X and LCM is Y" problems, since it reduces the search to just finding coprime m, n with m×n = Y/h.

4. Recognizing Word Problems

The hardest part of this chapter is rarely the calculation — it's recognizing which operation a word problem is actually asking for. Two phrasing patterns cover almost every case.

HCF phrasing — "largest/greatest that divides/fits evenly"
"Find the largest tile size that can exactly tile a floor
 594cm by 396cm, with no cutting."
 -> HCF of 594 and 396

"Three ropes of length 45m, 63m and 81m are to be cut into
 equal pieces, as long as possible, with none left over."
 -> HCF of 45, 63 and 81
LCM phrasing — "smallest/first time that coincides/repeats"
"Three bells ring every 12, 18 and 24 minutes. If they ring
 together now, when will they next ring together?"
 -> LCM of 12, 18 and 24

"Find the smallest number that leaves remainder 0 when
 divided by 4, 6 and 9."
 -> LCM of 4, 6 and 9

The tell: HCF problems involve splitting something into the largest possible equal pieces; LCM problems involve events repeating and asking when they next align. A variant worth knowing separately: "smallest number that leaves remainder r when divided by each of several numbers" is LCM + r, not just LCM — the remainder shifts the whole answer by a constant.

5. Practice Problems

Practice

30 problems, no calculator — Easy, Medium & Tough

Identify HCF vs. LCM from the phrasing before you start calculating. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.

Easy (1–10)

  1. Find the prime factorization of 84.
  2. Find the HCF of 12 and 18.
  3. Find the LCM of 4 and 6.
  4. Find the HCF of 24 and 36.
  5. Find the LCM of 5 and 8.
  6. Is 51 a prime number?
  7. Find the HCF of 15, 20 and 25.
  8. Find the LCM of 3, 4 and 5.
  9. Find the number of factors of 12.
  10. Find the HCF of 100 and 75.
Easy — Answers

1) 84 = 2² × 3 × 7. 2) HCF(12,18) = 6. 3) LCM(4,6) = 12. 4) HCF(24,36) = 12. 5) LCM(5,8) = 40 (coprime, so LCM = product). 6) No — 51 = 3 × 17. 7) HCF(15,20,25) = 5. 8) LCM(3,4,5) = 60 (all coprime, so LCM = product). 9) Factors of 12: 1,2,3,4,6,12 → 6 factors. 10) HCF(100,75) = 25.

Medium (11–20)

  1. Two numbers have HCF 6 and product 216. Find their LCM.
  2. Find the HCF and LCM of 48 and 60.
  3. Find the largest number that divides both 245 and 1,029 exactly.
  4. Three bells ring at intervals of 4, 6 and 8 minutes. If they ring together now, after how many minutes will they next ring together?
  5. Find the smallest number exactly divisible by 12, 16 and 24.
  6. The HCF of two numbers is 12 and their LCM is 144. If one number is 36, find the other.
  7. Find the HCF of 84, 126 and 210.
  8. Find the LCM of 15, 20 and 30.
  9. Two numbers are in the ratio 3:4 and their LCM is 84. Find the numbers.
  10. Find the smallest number which, when divided by 6, 9 and 12, leaves no remainder.
Medium — Answers

11) LCM = 216/6 = 36. 12) 48=2⁴×3, 60=2²×3×5 → HCF=2²×3=12, LCM=2⁴×3×5=240. 13) 245=5×7², 1029=3×7³ → HCF=7²=49. 14) LCM(4,6,8)=24 minutes. 15) LCM(12,16,24)=48. 16) Other number = (12×144)/36 = 48. 17) 84=2²×3×7, 126=2×3²×7, 210=2×3×5×7 → HCF=2×3×7=42. 18) LCM(15,20,30)=60. 19) Numbers are 3x,4x (coprime factors) → LCM=12x=84 → x=7 → numbers are 21 and 28. 20) LCM(6,9,12)=36.

Tough (21–30)

  1. Find the smallest number that leaves remainder 3 when divided by 6, 8 and 12.
  2. The product of two numbers is 4,032 and their HCF is 12. Find the number of possible pairs of such numbers.
  3. Find the HCF of 2³×3²×5 and 2²×3³×7.
  4. Three numbers are in the ratio 2:3:4 and their HCF is 12. Find the numbers.
  5. Find the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively.
  6. Four bells toll at intervals of 6, 8, 12 and 18 seconds. They all toll together at 9:00:00 am. At what time will they next toll together?
  7. Find the number of pairs of positive integers (a,b) such that HCF(a,b)=4 and LCM(a,b)=48.
  8. The HCF of two numbers is 23, and the other two factors of their LCM are 13 and 14. Find the larger of the two numbers.
  9. Find the smallest 4-digit number which is exactly divisible by 12, 15, 18 and 27.
  10. Two tankers contain 850 liters and 680 liters of fuel. Find the maximum capacity of a container that can measure the fuel of both tankers an exact number of times.
Tough — Answers

21) LCM(6,8,12)=24; answer=24+3=27. 22) LCM=4032/12=336; 336/12=28=2²×7; coprime factor pairs of 28: (1,28) and (4,7) → 2 pairs. 23) HCF takes the smallest power of each common prime: 2²×3²=36. 24) HCF of 2x,3x,4x is x (since gcd(2,3,4)=1) → x=12 → numbers are 24, 36, 48. 25) 70-5=65, 125-8=117; HCF(65,117): 65=5×13, 117=9×13 → HCF=13. 26) LCM(6,8,12,18)=72 seconds = 1 min 12 sec → next toll at 9:01:12 am. 27) LCM/HCF=48/4=12=2²×3; coprime factor pairs of 12: (1,12) and (3,4) → 2 pairs. 28) Numbers = 23×13=299 and 23×14=322 → larger is 322. 29) LCM(12,15,18,27): 12=2²×3, 15=3×5, 18=2×3², 27=3³ → LCM=2²×3³×5=540; smallest 4-digit multiple = 540×2=1,080. 30) HCF(850,680): 850=2×5²×17, 680=2³×5×17 → HCF=2×5×17=170 liters.

6. Knowledge Check

Four quick questions. Expand each to check your answer.

Q1

Why does HCF take the SMALLEST power of each common prime, while LCM takes the LARGEST power of every prime appearing anywhere?

HCF must divide every original number, so it can only use as much of each prime as the number with the LEAST of that prime has. LCM must be divisible BY every original number, so it needs at least as much of each prime as the number with the MOST of that prime requires — the two definitions pull the exponent in opposite directions.

Q2

Why does the HCF × LCM = product identity work for exactly two numbers but not generally for three or more?

For two numbers, the min-exponent (HCF) and max-exponent (LCM) of each prime always add up to exactly the sum of the two original exponents, which makes the product identity hold exactly. With three or more numbers, the min and max no longer account for every number's exponent (a middle value can be ignored by both), so the simple multiplicative relationship breaks down.

Q3

What phrasing distinguishes an HCF word problem from an LCM word problem?

HCF problems describe splitting something into the largest possible equal pieces with nothing left over (tiling, cutting ropes). LCM problems describe repeating events and ask when they next align or coincide (bells ringing together, cycles repeating) — the "splitting large" vs. "repeating until aligned" framing is the reliable tell.

Q4

Why is "the smallest number leaving remainder 5 when divided by 8, 12 and 15" equal to LCM(8,12,15) + 5, not just LCM(8,12,15)?

The LCM itself is exactly divisible by all three numbers, leaving remainder 0, not 5. Adding the target remainder shifts that number up by exactly the right amount so that dividing by each of 8, 12 and 15 now leaves remainder 5 in every case, since the LCM portion still divides out evenly and only the added 5 remains.

← Back to Week 1: Number Systems, Divisibility & Remainders Up next Week 3: Percentages →