Highest common factor, least common multiple and their word problems.
45 solved questions · 18 easy · 17 medium · 10 hard
Take a 10-question quiz on HCF & LCM, no loginTry each one first, then open the answer and the step-by-step solution.
Find the HCF of 36 and 48.
Answer: A. 12
Solution
36 = 2^2 x 3^2 48 = 2^4 x 3 HCF = product of common prime powers (lowest powers) = 12. Answer: 12
Shortcut: Use the Euclidean method: divide the bigger by the smaller and repeat with remainders.
Find the HCF of 84 and 126.
Answer: C. 42
Solution
84 = 2^2 x 3 x 7 126 = 2 x 3^2 x 7 HCF = product of common prime powers (lowest powers) = 42. Answer: 42
Shortcut: Use the Euclidean method: divide the bigger by the smaller and repeat with remainders.
Find the HCF of 144 and 180.
Answer: B. 36
Solution
144 = 2^4 x 3^2 180 = 2^2 x 3^2 x 5 HCF = product of common prime powers (lowest powers) = 36. Answer: 36
Shortcut: Use the Euclidean method: divide the bigger by the smaller and repeat with remainders.
Find the HCF of 105 and 245.
Answer: D. 35
Solution
105 = 3 x 5 x 7 245 = 5 x 7^2 HCF = product of common prime powers (lowest powers) = 35. Answer: 35
Shortcut: Use the Euclidean method: divide the bigger by the smaller and repeat with remainders.
Find the LCM of 12 and 18.
Answer: D. 36
Solution
12 = 2^2 x 3 18 = 2 x 3^2 LCM = product of highest powers of all primes = 36. Answer: 36
Shortcut: LCM = (a x b) / HCF.
Find the LCM of 15 and 20.
Answer: B. 60
Solution
15 = 3 x 5 20 = 2^2 x 5 LCM = product of highest powers of all primes = 60. Answer: 60
Shortcut: LCM = (a x b) / HCF.
Find the LCM of 24 and 36.
Answer: A. 72
Solution
24 = 2^3 x 3 36 = 2^2 x 3^2 LCM = product of highest powers of all primes = 72. Answer: 72
Shortcut: LCM = (a x b) / HCF.
The HCF of two numbers is 12 and their LCM is 180. If one number is 36, what is the other?
Answer: C. 60
Solution
HCF x LCM = product of the numbers. 12 x 180 = 2160 = 36 x other. Other = 2160/36 = 60. Answer: 60
The product of two numbers is 1620 and their HCF is 9. What is their LCM?
Answer: C. 180
Solution
LCM = product / HCF = 1620 / 9 = 180. Answer: 180
Find the LCM of 8, 12 and 20.
Answer: A. 120
Solution
Prime factorisation: 8 = 2^3; 12 = 2^2 x 3; 20 = 2^2 x 5. Take the highest power of each prime. LCM = 120. Answer: 120
Find the LCM of 6, 9 and 15.
Answer: D. 90
Solution
Prime factorisation: 6 = 2 x 3; 9 = 3^2; 15 = 3 x 5. Take the highest power of each prime. LCM = 90. Answer: 90
Find the LCM of 4, 10 and 25.
Answer: B. 100
Solution
Prime factorisation: 4 = 2^2; 10 = 2 x 5; 25 = 5^2. Take the highest power of each prime. LCM = 100. Answer: 100
Find the HCF of 24, 60 and 84.
Answer: A. 12
Solution
Prime factorisation: 24 = 2^3 x 3; 60 = 2^2 x 3 x 5; 84 = 2^2 x 3 x 7. Common prime powers give HCF = 12. Answer: 12
Find the HCF of 42, 63 and 105.
Answer: C. 21
Solution
Prime factorisation: 42 = 2 x 3 x 7; 63 = 3^2 x 7; 105 = 3 x 5 x 7. Common prime powers give HCF = 21. Answer: 21
Which of the following pairs of numbers is co-prime?
Answer: B. (35, 18)
Solution
Co-prime means HCF = 1. (21, 28): HCF 7; (24, 36): HCF 12; (27, 45): HCF 9; (35, 18): HCF 1. Answer: (35, 18)
What is the LCM of two co-prime numbers 7 and 13?
Answer: D. 91
Solution
For co-prime numbers LCM = product = 7 x 13 = 91. Answer: 91
What is the HCF of 2^3 x 3^2 x 5 and 2^2 x 3^4 x 7?
Answer: D. 36
Solution
Take the lowest power of each common prime: 2^2 and 3^2. HCF = 4 x 9 = 36. Answer: 36
What is the LCM of 2^3 x 3^2 and 2^2 x 3^4?
Answer: B. 648
Solution
Take the highest power of each prime: 2^3 and 3^4. LCM = 8 x 81 = 648. Answer: 648
Three bells toll at intervals of 12, 15 and 20 minutes. If they toll together at 8:00 a.m., when will they next toll together?
Answer: A. 1 hour later, at 9:00 a.m.
Solution
They toll together after LCM of 12, 15, 20 units. LCM = 60. Answer: 60 minutes = 1 hour(s), so at 9:00 a.m.
Three traffic lights change after every 18, 24 and 30 seconds. If they change together at 6:00:00 a.m., after how many seconds will they change together again?
Answer: C. 360 seconds
Solution
They toll together after LCM of 18, 24, 30 units. LCM = 360. Answer: 360 seconds
Three ropes of lengths 120 m, 180 m and 300 m are to be cut into pieces of equal length, with no rope left over. What is the greatest possible length of each piece?
Answer: C. 60 m
Solution
The greatest equal length is the HCF of 120, 180 and 300. 120 = 2^3 x 3 x 5; 180 = 2^2 x 3^2 x 5; 300 = 2^2 x 3 x 5^2. HCF = 2^2 x 3 x 5 = 60. Answer: 60 m
A hall is 15 m 17 cm long and 9 m 2 cm broad. What is the side of the largest square tile that can pave the floor exactly without cutting?
Answer: A. 41 cm
Solution
Convert to cm: length = 1517 cm, breadth = 902 cm. Side = HCF(1517, 902). 1517 = 902 x 1 + 615; 902 = 615 x 1 + 287; 615 = 287 x 2 + 41; 287 = 41 x 7 + 0. HCF = 41. Answer: 41 cm
What is the smallest number which leaves remainder 3 when divided by 6, 8, 12 and 15?
Answer: D. 123
Solution
Smallest number divisible by all = LCM(6, 8, 12, 15) = 120. Add the common remainder 3: 120 + 3 = 123. Answer: 123
Find the smallest number which when divided by 12, 16, 18 and 30 leaves remainders 7, 11, 13 and 25 respectively.
Answer: B. 715
Solution
Divisor minus remainder is constant: 12-7 = 5, 16-11 = 5, 18-13 = 5, 30-25 = 5. So number = LCM - 5. LCM(12, 16, 18, 30) = 720. Number = 720 - 5 = 715. Answer: 715
Shortcut: If divisor - remainder is constant k, answer = LCM - k.
What is the greatest number which divides 1657 and 2037 leaving remainders 6 and 5 respectively?
Answer: A. 127
Solution
Subtract remainders: 1657 - 6 = 1651 and 2037 - 5 = 2032. Required number = HCF(1651, 2032). 2032 = 1651 x 1 + 381; 1651 = 381 x 4 + 127; 381 = 127 x 3 + 0. HCF = 127. Answer: 127
What is the greatest number which divides 1305, 4665 and 6905 leaving the same remainder in each case?
Answer: C. 1120
Solution
The number divides the differences: 4665 - 1305 = 3360 and 6905 - 4665 = 2240. HCF(3360, 2240): 3360 = 2240 x 1 + 1120; 2240 = 1120 x 2. HCF = 1120. Answer: 1120
Shortcut: Take the HCF of the differences between the numbers.
The LCM of two numbers is 2310 and their HCF is 30. If one number is 330, find the other.
Answer: B. 210
Solution
Other = (HCF x LCM)/330 = (30 x 2310)/330 = 69300/330 = 210. Answer: 210
Two numbers are in the ratio 3 : 4 and their LCM is 240. What is their HCF?
Answer: D. 20
Solution
Let the numbers be 3k and 4k, with HCF k. LCM of 3k and 4k = 12k = 240, so k = 20. The numbers are 60 and 80, HCF = 20. Answer: 20
The sum of two numbers is 80 and their HCF is 8. Which of the following could be the numbers?
Answer: D. 24 and 56
Solution
Both numbers must be multiples of 8 and add to 80, with the multipliers co-prime (otherwise the HCF is larger). 24 and 56 = 8 x 3 and 8 x 7, HCF = 8. 16 and 64: HCF 16. 32 and 48: HCF 16. 40 and 40: HCF 40. 20 and 60: not multiples of 8. Answer: 24 and 56
Find the HCF of 2/3, 4/9 and 10/27.
Answer: B. 2/27
Solution
HCF of fractions = HCF of numerators / LCM of denominators. HCF(2, 4, 10) = 2; LCM(3, 9, 27) = 27. HCF = 2/27. Answer: 2/27
Find the LCM of 2/3, 4/9 and 8/27.
Answer: A. 8/3
Solution
LCM of fractions (in lowest terms) = LCM of numerators / HCF of denominators. LCM(2, 4, 8) = 8; HCF(3, 9, 27) = 3. LCM = 8/3. Answer: 8/3
Two runners start together from the same point on a circular track and run in the same direction. One takes 12 minutes per lap and the other takes 18 minutes. After how many minutes will they next be together at the starting point?
Answer: C. 36
Solution
They meet at the start after LCM(12, 18) minutes. 12 = 2^2 x 3, 18 = 2 x 3^2, LCM = 4 x 9 = 36. Answer: 36 minutes
Three friends start walking round a park together. Their steps measure 40 cm, 60 cm and 90 cm. What is the minimum distance each must walk so that each covers the distance in a whole number of steps?
Answer: C. 3 m 60 cm
Solution
Minimum distance = LCM(40, 60, 90) cm. 40 = 2^3 x 5, 60 = 2^2 x 3 x 5, 90 = 2 x 3^2 x 5. LCM = 8 x 9 x 5 = 360 cm = 3 m 60 cm. Answer: 3 m 60 cm
What is the smallest 4-digit number exactly divisible by 12, 15, 20 and 35?
Answer: A. 1260
Solution
LCM(12, 15, 20, 35) = 2^2 x 3 x 5 x 7 = 420. 1000 / 420 = 2 remainder 160, so the next multiple of 420 above 1000 is 420 x 3 = 1260. Answer: 1260
What is the greatest 4-digit number exactly divisible by 24, 36 and 40?
Answer: D. 9720
Solution
LCM(24, 36, 40) = 2^3 x 3^2 x 5 = 360. 9999 / 360 = 27 remainder 279. Greatest = 9999 - 279 = 9720. Answer: 9720
The HCF of two numbers is 12 and their LCM is 336. How many such pairs of numbers exist?
Answer: B. 2
Solution
Let the numbers be 12m and 12n with m, n co-prime. LCM = 12mn = 336, so mn = 28. Co-prime pairs with product 28: (1, 28) and (4, 7). Numbers: (12, 336) and (48, 84). Answer: 2
What is the smallest perfect square exactly divisible by 12, 18 and 30?
Answer: A. 900
Solution
LCM(12, 18, 30) = 2^2 x 3^2 x 5 = 180. For a perfect square every prime needs an even power: 2^2 x 3^2 x 5^2 = 900. Answer: 900
What is the smallest perfect cube exactly divisible by both 12 and 18?
Answer: C. 216
Solution
LCM(12, 18) = 36 = 2^2 x 3^2. For a perfect cube every prime power must be a multiple of 3: 2^3 x 3^3 = 216. Answer: 216
How many numbers from 1 to 500 are divisible by each of 6, 8 and 9?
Answer: B. 6
Solution
Divisible by 6, 8 and 9 means divisible by LCM(6, 8, 9) = 72. Multiples of 72 up to 500: floor(500/72) = 6. Answer: 6
What is the smallest number which when divided by 5, 6, 7 and 8 leaves remainders 4, 5, 6 and 7 respectively?
Answer: D. 839
Solution
Each remainder is 1 less than its divisor, so number = LCM - 1. LCM(5, 6, 7, 8) = 840. Number = 839. Answer: 839
What is the smallest number which leaves remainder 1 when divided by 4, 5 and 6 but is exactly divisible by 7?
Answer: D. 301
Solution
Number = 60k + 1 (LCM(4, 5, 6) = 60), and it must be a multiple of 7. 60k + 1 mod 7: 60 = 7 x 8 + 4, so need 4k + 1 = 0 mod 7, i.e. 4k = 6 mod 7, k = 5 (4 x 5 = 20 = 6 mod 7). Number = 60 x 5 + 1 = 301 = 7 x 43. Answer: 301
Three numbers are in the ratio 2 : 3 : 5 and their HCF is 12. What is their LCM?
Answer: B. 360
Solution
Numbers = 12 x 2, 12 x 3, 12 x 5 = 24, 36, 60. LCM(24, 36, 60): 24 = 2^3 x 3, 36 = 2^2 x 3^2, 60 = 2^2 x 3 x 5. LCM = 8 x 9 x 5 = 360. Answer: 360
A rectangular floor measures 1517 cm by 902 cm. It is to be covered with identical square tiles of the largest possible size, none cut. How many tiles are needed?
Answer: A. 814
Solution
Largest tile side = HCF(1517, 902) = 41 cm. Tiles along length = 1517/41 = 37; along breadth = 902/41 = 22. Total = 37 x 22 = 814. Answer: 814
The LCM of two numbers is 14 times their HCF, and the sum of their HCF and LCM is 600. If one number is 280, what is the other?
Answer: C. 80
Solution
Let HCF = h, so LCM = 14h and h + 14h = 600, giving h = 40 and LCM = 560. Product of numbers = HCF x LCM = 40 x 560 = 22400. Other = 22400 / 280 = 80. Answer: 80
The HCF of two numbers is 16 and the other two factors of their LCM are 11 and 13. What is the larger of the two numbers?
Answer: C. 208
Solution
LCM = HCF x 11 x 13, so the two numbers are 16 x 11 = 176 and 16 x 13 = 208. Check: HCF(176, 208) = 16 and LCM = 2288. Larger = 208. Answer: 208
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