HCF and LCM
CSAT · Quantitative Aptitude
Introduction
HCF and LCM questions are quick marks in CSAT. They also appear disguised: bells ringing together, traffic lights, tiles that fit a floor, or the largest container that measures several quantities exactly. Recognising which of the two is needed is half the work.
Key concepts
Definitions
- HCF (highest common factor) – the largest number that divides all the given numbers exactly. Use it when you split things into the largest equal pieces.
- LCM (least common multiple) – the smallest number that all the given numbers divide exactly. Use it when events repeat and you want them to coincide.
Methods
- Prime factorisation – HCF = product of the common primes with the lowest powers; LCM = product of all primes with the highest powers. Example: 72 = 2³·3², 120 = 2³·3·5 → HCF = 2³·3 = 24, LCM = 2³·3²·5 = 360.
- Division method (Euclid) – HCF(a, b) = HCF(b, a mod b). HCF(391, 323): 391 mod 323 = 68; 323 mod 68 = 51; 68 mod 51 = 17; 51 mod 17 = 0 → 17.
Fractions
HCF of fractions = HCF of numerators ÷ LCM of denominators. LCM of fractions = LCM of numerators ÷ HCF of denominators.
Word-problem patterns
| Wording | Use |
|---|---|
| Largest tile / container / group size | HCF |
| Bells, lights, runners meeting again | LCM |
| Greatest number dividing a, b, c leaving the same remainder | HCF of the differences |
| Smallest number leaving remainder r with each divisor | LCM + r |
| Greatest number leaving remainders r1, r2 … | HCF of (a − r1), (b − r2) … |
- HCF × LCM = product of the two numbers (true for two numbers only)
- HCF always divides the LCM
- Same remainder: HCF(|a − b|, |b − c|)
- Remainder r with each divisor: k × LCM + r
Solved examples
Example 1
Find the HCF of 6, 33 and 9.
(a) 6 (b) 19 (c) 3 (d) 1
- 6 = 3×2, 33 = 3×11, 9 = 3×3.
- 2, 11, 3 have no common factor, so HCF = 3.
Answer: (c) 3
Example 2
Three bells toll at intervals of 6, 9 and 15 seconds. They toll together at 12:00. How many times will they toll together in 1 hour, including the toll at 12:00?
(a) 41 (b) 40 (c) 42 (d) 240
- They toll together every LCM(6, 9, 15) = 90 seconds.
- In 3600 seconds that is 3600 ÷ 90 = 40 intervals.
- Add the first toll: 41.
Answer: (a) 41
Example 3
What is the greatest number that divides 159, 250 and 640 leaving the same remainder in each case?
(a) 1 (b) 13 (c) 3 (d) 26
- If all leave the same remainder, the differences are exact multiples of the divisor.
- Differences: 91 and 390.
- HCF(91, 390) = 13.
Answer: (b) 13
Example 4
What is the smallest number greater than 1 which leaves a remainder of 1 when divided by 10, 12 and 15?
(a) 121 (b) 60 (c) 59 (d) 61
- LCM(10, 12, 15) = 60.
- Add the remainder: 60 + 1 = 61.
Answer: (d) 61
Example 5
The HCF of two numbers is 15 and their LCM is 495. If one number is 45, what is the other?
(a) 337 (b) 165 (c) 180 (d) 33
- Product of two numbers = HCF × LCM.
- Other number = 15 × 495 ÷ 45 = 7425 ÷ 45 = 165.
Answer: (b) 165
Example 6
What is the smallest number greater than 4 which leaves a remainder of 4 when divided by 5, 6 and 12?
(a) 64 (b) 60 (c) 124 (d) 56
- LCM(5, 6, 12) = 60.
- Add the remainder: 60 + 4 = 64.
Answer: (a) 64
Example 7
What is the greatest number that divides 222, 625 and 651 leaving the same remainder in each case?
(a) 14 (b) 13 (c) 26 (d) 1
- If all leave the same remainder, the differences are exact multiples of the divisor.
- Differences: 403 and 26.
- HCF(403, 26) = 13.
Answer: (b) 13
Example 8
The HCF of two numbers is 12 and their LCM is 780. If one number is 156, what is the other?
(a) 60 (b) 65 (c) 72 (d) 936
- Product of two numbers = HCF × LCM.
- Other number = 12 × 780 ÷ 156 = 9360 ÷ 156 = 60.
Answer: (a) 60
- If one number divides the other, the smaller is the HCF and the larger the LCM.
- The HCF must divide every difference between the numbers – test options that way.
- For 'how many times together in an hour', divide 3,600 s by the LCM and add 1 for the starting ring.
- Using HCF × LCM = product for three numbers – it does not hold.
- Forgetting to add the first coincidence at time zero.
- Mixing units (minutes and seconds) before taking an LCM.
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. What is the smallest number greater than 7 which leaves a remainder of 7 when divided by 8, 9 and 18?
(a) 151 (b) 65 (c) 79 (d) 72
Q2. What is the greatest number that divides 211, 224 and 744 leaving the same remainder in each case?
(a) 3 (b) 13 (c) 26 (d) 1
Q3. What is the greatest number that divides 73, 260 and 326 leaving the same remainder in each case?
(a) 11 (b) 1 (c) 22 (d) 7
Q4. What is the greatest number that divides 87, 538 and 648 leaving the same remainder in each case?
(a) 22 (b) 11 (c) 10 (d) 1
Q5. Three bells toll at intervals of 8, 12 and 24 seconds. They toll together at 12:00. How many times will they toll together in 3 hours, including the toll at 12:00?
(a) 452 (b) 496 (c) 450 (d) 451
Q6. Three lights blink every 6, 9 and 24 seconds and blink together at a moment. After how many seconds will they next blink together?
(a) 39 (b) 144 (c) 432 (d) 72
Q7. Find the LCM of 4, 22 and 6.
(a) 264 (b) 145 (c) 132 (d) 66
Q8. What is the smallest number greater than 3 which leaves a remainder of 3 when divided by 5, 8 and 9?
(a) 357 (b) 360 (c) 363 (d) 723
Q9. What is the greatest number that divides 90, 376 and 701 leaving the same remainder in each case?
(a) 26 (b) 13 (c) 1 (d) 12
Q10. Find the LCM of 35, 15 and 10.
(a) 210 (b) 1050 (c) 420 (d) 30
Q11. The HCF of two numbers is 16 and their LCM is 2288. If one number is 208, what is the other?
(a) 1664 (b) 143 (c) 192 (d) 176
Q12. What is the smallest number greater than 4 which leaves a remainder of 4 when divided by 5, 8 and 12?
(a) 124 (b) 116 (c) 120 (d) 244
Q1 – (c) 79. LCM(8, 9, 18) = 72. Add the remainder: 72 + 7 = 79.
Q2 – (b) 13. If all leave the same remainder, the differences are exact multiples of the divisor. Differences: 13 and 520. HCF(13, 520) = 13.
Q3 – (a) 11. If all leave the same remainder, the differences are exact multiples of the divisor. Differences: 187 and 66. HCF(187, 66) = 11.
Q4 – (b) 11. If all leave the same remainder, the differences are exact multiples of the divisor. Differences: 451 and 110. HCF(451, 110) = 11.
Q5 – (d) 451. They toll together every LCM(8, 12, 24) = 24 seconds. In 10800 seconds that is 10800 ÷ 24 = 450 intervals. Add the first toll: 451.
Q6 – (d) 72. They blink together again after LCM(6, 9, 24) = 72 seconds.
Q7 – (c) 132. Write each as 2 × a prime: 4, 22, 6. LCM = 2 × 2 × 11 × 3 = 132.
Q8 – (c) 363. LCM(5, 8, 9) = 360. Add the remainder: 360 + 3 = 363.
Q9 – (b) 13. If all leave the same remainder, the differences are exact multiples of the divisor. Differences: 286 and 325. HCF(286, 325) = 13.
Q10 – (a) 210. Write each as 5 × a prime: 35, 15, 10. LCM = 5 × 7 × 3 × 2 = 210.
Q11 – (d) 176. Product of two numbers = HCF × LCM. Other number = 16 × 2288 ÷ 208 = 36608 ÷ 208 = 176.
Q12 – (a) 124. LCM(5, 8, 12) = 120. Add the remainder: 120 + 4 = 124.
- HCF = split into largest equal parts; LCM = events coincide
- HCF × LCM = a × b (two numbers)
- Same remainder → HCF of differences
- Given remainder r → LCM + r
