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HCF & LCM

Mathematics · Arithmetic Foundations

Useful for: SSC CGL, Railway RRB, OSSC RHT — Regular High School Teacher, OSEPA Primary Teacher / JT Category-I, OSEPA Upper Primary — Science, OSSC Mathematics, OSSSC Govt Physical Education Teacher (PET), OSSSC Govt TGT Arts

and 4 more exams

OSSSC Govt TGT Science — CBZ, OSSSC Govt TGT Science — PCM, SSB Lecturer (Degree Colleges), SSB TGT PCM

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Overview

HCF is the largest common factor, LCM the smallest common multiple; remainder questions are HCF or LCM problems in disguise.

Key points

  • HCF × LCM = product of the two numbers (only for two numbers).
  • HCF of fractions = HCF of numerators ÷ LCM of denominators; LCM of fractions = LCM of numerators ÷ HCF of denominators.
  • Largest number dividing a, b, c leaving the same remainder r: HCF(a − r, b − r, c − r).
  • Largest number dividing a, b, c leaving the same unknown remainder: HCF of their differences.
  • Least number which leaves remainder r when divided by x, y, z: LCM(x, y, z) + r.

Explanation

  • When remainders differ but (divisor − remainder) is constant k: answer = LCM − k.
  • Bells ringing together, runners meeting at the start, traffic lights: these are LCM problems.
  • Cutting into the largest equal pieces or the fewest equal groups: HCF problems.
  • Remainder theorem: remainder of (a × b) ÷ n = (remainder of a × remainder of b) ÷ n.

Step by step

  1. Decide HCF or LCM from the wording.
  2. Adjust for remainders.
  3. Compute and verify with division.

Examples

  • HCF = 12, LCM = 360, one number = 72 → other = 12 × 360 ÷ 72 = 60.
  • Largest number dividing 245 and 1029 leaving remainder 5 each: HCF(240, 1024) = 16.
  • Least number which leaves remainder 4 when divided by 12, 15 and 20: LCM 60 + 4 = 64.

Traps and advanced points

  • Trap: Using HCF × LCM = product for three numbers — it does not hold.
  • Trap: Adding the remainder in an HCF question instead of subtracting it first.
  • Trap: Confusing "largest number that divides" (HCF) with "least number divisible by" (LCM).

Quick revision

  • HCF divides LCM always.
  • Same remainder → subtract first, then HCF.
  • Bells/runners → LCM.

Practice MCQs

  1. Find the LCM of 30 and 75.

    1. 150
    2. 180
    3. 165
    4. 135

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    Answer: (a) 150. LCM × HCF = product of the two numbers.

  2. Find the LCM of 36 and 48.

    1. 144
    2. 168
    3. 156
    4. 132

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    Answer: (a) 144. LCM × HCF = product of the two numbers.

  3. Find the LCM of 44 and 66.

    1. 110
    2. 176
    3. 154
    4. 132

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    Answer: (d) 132. LCM × HCF = product of the two numbers.

  4. Find the LCM of 52 and 52.

    1. 156
    2. 104
    3. 52
    4. 0

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    Answer: (c) 52. LCM × HCF = product of the two numbers.

  5. Find the LCM of 60 and 72.

    1. 360
    2. 384
    3. 348
    4. 372

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    Answer: (a) 360. LCM × HCF = product of the two numbers.

  6. The HCF of two numbers is 12 and their LCM is 336. If one number is 84, the other is:

    1. 36
    2. 96
    3. 72
    4. 48

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    Answer: (d) 48. Rule: HCF × LCM = product of two numbers.

  7. Find the greatest number that divides 43, 91 and 183 leaving the same remainder in each case.

    1. 13
    2. 4
    3. 7
    4. 9

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    Answer: (b) 4. Same remainder wala → differences ka HCF.

  8. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case is:

    1. 843
    2. 840
    3. 837
    4. 823

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    Answer: (a) 843. LCM(5,6,7,8) = 840.

  9. Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 8:00 AM, they will next ring together at:

    1. 10:00 AM
    2. 9:30 AM
    3. 12:00 PM
    4. 11:00 AM

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    Answer: (d) 11:00 AM. Saath bajne ka time = LCM(9,12,15).

  10. The HCF of 2^3 × 3^2 × 5 and 2^2 × 3^3 × 7 is:

    1. 2^2 × 3^3
    2. 2^3 × 3^2 × 5
    3. 2^2 × 3^2
    4. 2^3 × 3^3

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    Answer: (c) 2^2 × 3^2. HCF = common prime factors ki lowest power.

  11. Two numbers are in the ratio 3:4 and their LCM is 240. The smaller number is:

    1. 80
    2. 48
    3. 36
    4. 60

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    Answer: (d) 60. Numbers = 3k, 4k.

  12. The greatest number of 4 digits which is exactly divisible by 12, 15 and 18 is:

    1. 9960
    2. 9900
    3. 9980
    4. 9720

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    Answer: (b) 9900. LCM(12,15,18)=180.

  13. The LCM of two co-prime numbers is 143. If one number is 11, the other is:

    1. 33
    2. 11
    3. 13
    4. 12

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    Answer: (c) 13. Co-prime numbers ka HCF = 1, isliye LCM = product = 143.

  14. Find the least number which when increased by 7 is divisible by 8, 10 and 12.

    1. 113
    2. 120
    3. 127
    4. 233

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    Answer: (a) 113. LCM(8,10,12)=120.

  15. The product of two numbers is 2028 and their HCF is 13. The number of such pairs is:

    1. 2
    2. 3
    3. 4
    4. 1

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    Answer: (a) 2. Numbers = 13a, 13b (a,b co-prime).

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