HCF and LCM Word Problems: Remainders, Bells, Tracks and Tiles
HCF and LCM Word Problems: Remainders, Bells, Tracks and Tiles
Most NTPC HCF-LCM questions are word problems in disguise. The decision is always the same: is the question looking for the GREATEST number that divides (HCF) or the LEAST number that is divisible (LCM)? Then adjust for the remainder. The table below covers every variant.
1. Master Decision Table
| Question asks for | Formula | Example numbers |
|---|---|---|
| Greatest number dividing a, b, c exactly | HCF(a, b, c) | 36, 84, 120 → 12 |
| Greatest number dividing a, b, c leaving the SAME remainder r | HCF(a − r, b − r, c − r) | |
| Greatest number dividing a, b, c leaving remainders r₁, r₂, r₃ | HCF(a − r₁, b − r₂, c − r₃) | 1657, 2037 with 6, 5 → HCF(1651, 2032) = 127 |
| Greatest number leaving the same (unknown) remainder | HCF of the differences | 445, 572, 699 → HCF(127, 127) = 127 |
| Least number divisible by a, b, c | LCM(a, b, c) | 12, 15, 20 → 60 |
| Least number leaving the SAME remainder r | LCM + r (the trivial answer r itself, with every quotient 0, is excluded by convention) | |
| Least number leaving remainders where (divisor − remainder) = k for all | LCM − k | 5, 6, 7, 8 leaving 3, 4, 5, 6 → 840 − 2 = 838 |
| Least such number that is also divisible by m | Test LCM × t ± adjustment for t = 1, 2, 3, ... | |
| Greatest n-digit number divisible by a, b, c | Largest n-digit number − (its remainder by LCM) | |
| Smallest n-digit number leaving remainder r | Smallest number of the form LCM × t + r that has n digits: try the multiple just below 10^(n − 1) plus r first, then the next multiple | 3-digit, remainder 3 by 9 and 11: 99 + 3 = 102 (not 198 + 3) |
| Bells, lights, alarms together | LCM of intervals | |
| Runners meet at the start | LCM of lap times | |
| Largest tile, can, rope piece, group size | HCF | |
| Least number of square tiles | (Length ÷ HCF) × (Breadth ÷ HCF) = Area ÷ HCF² | 624 cm × 432 cm, HCF 48 → 13 × 9 = 117 |
| Least number of cans | Total quantity ÷ HCF | 403 + 434 + 465 litres, HCF 31 → 42 cans |
| Count of times together in a period | ⌊Period/LCM⌋ + 1 (counting the start) |
✗ Least number leaving remainder 8 when divided by 12, 15, 20 = LCM − 8 | ✓ It is LCM + 8; "LCM − k" applies only when each divisor minus its remainder gives the same k
हिंदी नोट: जहाँ "सबसे बड़ी संख्या जो विभाजित करे" पूछा जाए वहाँ म.स. लगाइए, और जहाँ "सबसे छोटी संख्या जो विभाज्य हो" पूछा जाए वहाँ ल.स.। घंटियाँ एक साथ बजना और दौड़ में प्रारंभिक बिंदु पर मिलना हमेशा ल.स. से हल होता है।
Exam Pointer: Verified NTPC patterns: greatest number with given remainders (February 2021, June 2025 Graduate CBT-1), least number with remainder conditions (two March 2021 shifts), smallest six-digit common multiple (January 2021), traffic lights changing together (February 2021), runners on a circular track (January 2021) and HCF grouping of students (June 2022 CBT-2); counting common multiples below a limit is supported at the 2020-21 cycle level. Time each one: 40 seconds is the target.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Greatest number dividing with given or equal remainders
[PYQ: NTPC Graduate CBT-1 14-Jun-2025 Shift-2 | NTPC CBT-1 11-Feb-2021 Shift-2]
EXAM LEVEL
Q. Find the greatest number that divides 1,657 and 2,037 leaving remainders 6 and 5 respectively.
Subtract the remainders: 1,651 and 2,032. HCF by division: 2,032 − 1,651 = 381; 1,651 = 4 × 381 + 127; 381 = 3 × 127. HCF = 127.
Answer: 127
EXAMATLAS LEVEL
Q. Find the greatest number that divides 445, 572 and 699 leaving the same remainder in each case, and find that remainder.
The remainder is unknown, so use differences: 572 − 445 = 127 and 699 − 572 = 127. HCF = 127. Remainder = 445 − 3 × 127 = 445 − 381 = 64 (572 and 699 also leave 64).
Answer: 127, remainder 64
Pattern 2: Least number with remainder conditions
[PYQ: NTPC CBT-1 27-Mar-2021 Shift-2 | NTPC CBT-1 19-Mar-2021 Shift-2]
EXAM LEVEL
Q. Find the least number which, when divided by 12, 15, 20 and 54, leaves remainder 8 in each case.
LCM(12, 15, 20, 54) = 2² × 3³ × 5 = 540. Same remainder, so add it: 548. (The number 8 itself also leaves remainder 8 with quotient 0 each time; exam keys exclude that trivial case, which is why the answer is LCM + r.)
Answer: 548
EXAMATLAS LEVEL
Q. Find the least number which leaves remainders 3, 4, 5 and 6 when divided by 5, 6, 7 and 8 respectively, but is exactly divisible by 11.
Each divisor minus its remainder is 2, so N = 840t − 2 where 840 = LCM(5, 6, 7, 8). Now 840 leaves 4 on division by 11, so 840t − 2 leaves 4t − 2. Make it 0: 4t − 2 must be a multiple of 11; t = 6 gives 22. N = 840 × 6 − 2 = 5,038, and 5,038 = 11 × 458. Testing only t = 1 (838) is where most attempts stop.
Answer: 5,038
Pattern 3: Largest or smallest n-digit number with LCM conditions
[PYQ: NTPC CBT-1 8-Jan-2021 Shift-2]
EXAM LEVEL
Q. Find the largest four-digit number exactly divisible by 12, 15, 18 and 27.
LCM = 2² × 3³ × 5 = 540. 9,999 ÷ 540 gives quotient 18, so the largest multiple is 540 × 18 = 9,720.
Answer: 9,720
EXAMATLAS LEVEL
Q. Find the smallest five-digit number which, when divided by 16, 24, 30 and 36, leaves remainder 10 in each case.
LCM = 2⁴ × 3² × 5 = 720. Smallest five-digit multiple: 10,000 ÷ 720 gives 13.9, so take 720 × 14 = 10,080. Add the remainder: 10,090. Check that 720 × 13 + 10 = 9,370 is only four digits.
Answer: 10,090
Pattern 4: Counting numbers divisible by several numbers
[PYQ: NTPC 2020-21 cycle]
EXAM LEVEL
Q. How many numbers from 1 to 1,000 are divisible by 6, 8 and 12?
Divisible by all three means divisible by LCM = 24. ⌊1000/24⌋ = 41.
Answer: 41
EXAMATLAS LEVEL
Q. How many four-digit numbers are divisible by 12, 18 and 30?
LCM = 180. Multiples up to 9,999: ⌊9999/180⌋ = 55. Multiples up to 999: ⌊999/180⌋ = 5. Four-digit multiples = 55 − 5 = 50.
Answer: 50
Pattern 5: Bells, alarms and traffic lights together
[PYQ: NTPC CBT-1 16-Feb-2021 Shift-2]
EXAM LEVEL
Q. Four bells toll at intervals of 6, 8, 12 and 18 seconds. They toll together at the start. How many times do they toll together in 30 minutes, including the start?
LCM(6, 8, 12, 18) = 72 seconds. 30 minutes = 1,800 seconds and 1,800/72 = 25 more times after the start. Total = 25 + 1 = 26.
Answer: 26
EXAMATLAS LEVEL
Q. Traffic lights at three crossings change after every 48, 72 and 108 seconds. If they change together at 7:20:00 a.m., at what time will they next change together?
48 = 2⁴ × 3, 72 = 2³ × 3², 108 = 2² × 3³. LCM = 2⁴ × 3³ = 432 seconds = 7 minutes 12 seconds. Next change together at 7:27:12 a.m.
Answer: 7:27:12 a.m.
Pattern 6: Runners meeting at the starting point
[PYQ: NTPC CBT-1 7-Jan-2021 Shift-2]
EXAM LEVEL
Q. A, B and C start together from the same point of a circular track and complete one round in 36, 48 and 60 seconds. After how much time will they first meet at the starting point?
LCM(36, 48, 60) = 720 seconds = 12 minutes.
Answer: 12 minutes
EXAMATLAS LEVEL
Q. A, B and C complete one round of a park in 2 1/2, 3 1/3 and 4 minutes respectively. When will they first be together at the starting point, and how many rounds will A have completed by then?
Lap times are 5/2, 10/3 and 4/1. LCM of fractions = LCM(5, 10, 4)/HCF(2, 3, 1) = 20/1 = 20 minutes. A completes 20 ÷ 5/2 = 8 rounds (B completes 6, C completes 5, all whole numbers, confirming the answer).
Answer: 20 minutes; A completes 8 rounds
Pattern 7: Tiles, cans and equal groups (HCF applications)
[PYQ: NTPC CBT-2 16-Jun-2022 Shift-2]
EXAM LEVEL
Q. A room is 6 m 24 cm long and 4 m 32 cm wide. Find the least number of identical square tiles needed to cover the floor exactly.
Largest tile side = HCF(624, 432) cm = 48 cm. Number of tiles = (624/48) × (432/48) = 13 × 9 = 117.
Answer: 117
EXAMATLAS LEVEL
Q. Three tankers carry 403 L, 434 L and 465 L of milk. Find the capacity of the largest can that measures each exactly, and the total number of such cans needed.
HCF must divide the differences 31 and 31, and 31 divides all three (403 = 13 × 31, 434 = 14 × 31, 465 = 15 × 31). Capacity = 31 L. Cans = 13 + 14 + 15 = 42.
Answer: 31 L, 42 cans
60-Second Revision
- "Greatest that divides" means HCF; "least that is divisible" means LCM.
- Same remainder r: HCF(a − r, ...) or LCM + r; unknown equal remainder: HCF of differences.
- Divisor minus remainder constant k: LCM − k; extra divisibility condition: test multiples of LCM.
- Times together in a period = ⌊period/LCM⌋ + 1 with the start.
- Fractional lap times: LCM(num)/HCF(den).
- HCF gives the size; cans = total ÷ HCF; square tiles = (L ÷ HCF) × (B ÷ HCF).
Next Step: LCM and HCF done. Solve the LCM-HCF questions in the ExamAtlas RRB NTPC 2026 mock tests next, and time yourself at 40 seconds per question.