Prelims · 8 Mental Ability · 8.1 Number system and simplification

8.1.3 HCF and LCM: methods and word problems

HCF is the largest number that divides all the given numbers, LCM is the smallest number they all divide, and most word problems only ask which of the two is needed and how the remainder is adjusted.

  • APPSC Group 2 · Prelims
  • Standard level
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Core facts

  • HCF (highest common factor, also GCD): the greatest number that divides each given number exactly.
  • LCM (least common multiple): the smallest number exactly divisible by each given number.
  • Prime-factor method: HCF = common primes with the lowest powers; LCM = all primes with the highest powers.
  • Division (Euclid) method: divide the larger by the smaller, then the divisor by the remainder, until the remainder is 0. The last divisor is the HCF.
  • For two numbers only: HCF × LCM = product of the numbers. HCF always divides the LCM.
  • Fractions (in lowest terms): HCF = HCF of numerators ÷ LCM of denominators; LCM = LCM of numerators ÷ HCF of denominators.
  • Decimals: give all numbers the same number of decimal places, work with the whole numbers, then put the decimal places back.
  • Key words: "greatest", "largest size", "maximum number of equal groups" → HCF. "Least", "together again", "first time all" → LCM.

Example 1. Find the HCF and LCM of 72 and 120. Solution. 72 = 2³ × 3²; 120 = 2³ × 3 × 5. HCF = 2³ × 3 = 24; LCM = 2³ × 3² × 5 = 360. Answer: 24 and 360

Example 2. The product of two numbers is 4,320 and their LCM is 360. Find the HCF. Solution. HCF = 4,320 ÷ 360. Answer: 12

Example 3. Find the HCF of 493 and 899 by division. Solution. 899 = 1 × 493 + 406; 493 = 1 × 406 + 87; 406 = 4 × 87 + 58; 87 = 1 × 58 + 29; 58 = 2 × 29. Answer: 29

Explained

Type 1: ratio with HCF or LCM. Write the numbers as ak and bk with a, b co-prime; then HCF = k and LCM = abk.

Example 4. Two numbers are in the ratio 4 : 5 and their LCM is 420. Find their sum. (a) 168 (b) 189 (c) 210 (d) 231 Solution. LCM = 20k = 420, so k = 21. Numbers 84 and 105. Answer: (b) 189

Type 2: least number leaving the same remainder. Answer = LCM × k + remainder (k = 1 for the least).

Example 5. Least number that leaves 5 when divided by 8, 12, 18 and 30. Solution. LCM(8, 12, 18, 30) = 2³ × 3² × 5 = 360. 360 + 5. Answer: 365

Type 3: greatest number leaving given remainders. Subtract each remainder, then take the HCF.

Example 6. Greatest number that divides 1,342 and 2,058 leaving remainders 7 and 11. Solution. 1,342 − 7 = 1,335 = 89 × 15; 2,058 − 11 = 2,047 = 89 × 23. HCF = 89, which is larger than both remainders. Answer: 89

Type 4: greatest number leaving the same (unknown) remainder. Take the HCF of the differences.

Example 7. Greatest number that divides 1,277, 3,167 and 6,002 leaving the same remainder each time. Solution. Differences: 1,890, 2,835 and 4,725. HCF = 945. (Each number leaves 332.) Answer: 945

Type 5: tiles, rods and equal groups (HCF).

Example 8. A floor 6.24 m by 4.68 m is to be covered with the largest possible square tiles, no cutting. How many tiles are needed? Solution. In cm: HCF(624, 468) = 156, so tiles are 1.56 m square. Tiles = (624 ÷ 156) × (468 ÷ 156) = 4 × 3. Answer: 12

Type 6: bells, lights and runners (LCM).

Example 9. Three traffic lights change every 40, 56 and 84 seconds. They change together at 7:45:00 a.m. When do they next change together? Solution. LCM(40, 56, 84) = 2³ × 3 × 5 × 7 = 840 seconds = 14 min. Answer: 7:59:00 a.m.

Example 10. Four bells ring every 9, 12, 15 and 18 minutes and ring together at noon. How many times do they ring together in the 6 hours from noon, counting noon? Solution. LCM = 180 minutes. 360 ÷ 180 = 2 intervals, so 2 + 1 times. Answer: 3

Speed tips.

  • Decide HCF or LCM from the key word before calculating.
  • "Greatest … remainders": subtract the remainders first. "Least … remainder": add it after.
  • In "least number" MCQs, test the smallest option against the last condition first.

Common traps

  • ✗ "Bells ring together 2 times in 6 hours." ✓ Count the start too: 3 when the question says "counting noon".
  • ✗ "HCF × LCM = product of three numbers." ✓ The rule holds only for two numbers.
  • ✗ "Greatest number leaving remainders 7 and 11: take HCF of 1,342 and 2,058." ✓ Subtract the remainders first: HCF of 1,335 and 2,047.
  • ✗ "Same remainder each time: take the HCF of the numbers." ✓ Take the HCF of their differences.
  • ✗ "LCM of fractions = LCM of numerators ÷ LCM of denominators." ✓ Divide by the HCF of denominators.
  • ✗ "Two numbers can have HCF 12 and LCM 90." ✓ The LCM must be a multiple of the HCF; no such pair exists.

Quick revision

  • HCF: lowest powers of common primes. LCM: highest powers of all primes.
  • Two numbers: HCF × LCM = product; HCF divides LCM.
  • Fractions: HCF = HCF(num)/LCM(den); LCM = LCM(num)/HCF(den).
  • Same remainder r: answer = LCM × k + r.
  • Constant shortfall d: LCM − d.
  • Greatest divisor with given remainders: HCF of (number − remainder).
  • Greatest divisor with the same unknown remainder: HCF of differences.
  • Tiles and equal groups: HCF. Bells and lights: LCM, add 1 for the start.
  • Pairs with HCF h, LCM l: co-prime pairs with product l/h.
Go deeper: Deeper (beyond what APPSC Group 2 usually asks)

Deeper Advanced

Constant shortfall. If (divisor − remainder) is the same for every divisor, answer = LCM − that shortfall.

Example 11. Least number that leaves 4, 5 and 6 when divided by 5, 6 and 7. Solution. Each remainder is 1 short of its divisor. LCM(5, 6, 7) − 1 = 210 − 1. Answer: 209

Example 12. Least number that leaves 1, 2 and 3 when divided by 4, 5 and 7. (a) 13 (b) 17 (c) 37 (d) 57 Solution. Shortfalls 3, 3 and 4 are not equal, so build step by step. Numbers leaving 3 on 7: 3, 10, 17, 24, … Keep those leaving 2 on 5: 17 is the first. 17 ÷ 4 leaves 1. Answer: (b) 17

Example 13. Least number that leaves 4 when divided by 6, 9, 15 and 18 but is exactly divisible by 7. Solution. Form 90k + 4 (LCM 90). 90 leaves 6 on 7, so 6k + 4 must be a multiple of 7: k = 4 gives 28. Number = 364 = 7 × 52. Answer: 364

Example 14. How many pairs of natural numbers have HCF 12 and LCM 360? Solution. Numbers 12a and 12b with a, b co-prime and ab = 360 ÷ 12 = 30. Co-prime pairs: (1, 30), (2, 15), (3, 10), (5, 6). Answer: 4

  • If the LCM is not a multiple of the HCF, no pair exists.

Example 15. (i) LCM of 3/4, 5/6 and 9/10. (ii) HCF of 4/5, 6/25 and 8/15. (iii) HCF of 0.48, 0.72 and 1.08. Solution. (i) LCM(3, 5, 9) ÷ HCF(4, 6, 10) = 45/2. (ii) HCF(4, 6, 8) ÷ LCM(5, 25, 15) = 2/75. (iii) HCF(48, 72, 108) = 12, so 0.12. Answer: (i) 45/2 (ii) 2/75 (iii) 0.12

  • Three numbers in a ratio. Numbers in the ratio 2 : 3 : 4 with HCF 12 are 24, 36, 48; LCM = 144.
  • Product rule fails for three numbers. HCF × LCM = product only for two numbers.

Sources: NCERT Mathematics textbooks, Classes 6 and 10 (HCF and LCM, Euclid's division lemma, fundamental theorem of arithmetic)

Last checked 2026-10-17.

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