Problems on H.C.F and L.C.M Aptitude

Problems on H.C.F and L.C.M are vital in understanding the fundamental principles of number theory. H.C.F (also known as GCD – Greatest Common Divisor) of two or more numbers is the largest number that divides all of them without leaving a remainder. L.C.M is the smallest number that is a multiple of each of the numbers. These concepts are essential for solving problems related to divisibility, fractions, ratios, and more in competitive exams.

1. Find the H.C.F of 36 and 48.

H.C.F of 36 and 48 can be found by prime factorization.
36 = 2² * 3², 48 = 2⁴ * 3.
H.C.F = 2² * 3 = 12.

2. What is the L.C.M of 12 and 15?

L.C.M of 12 and 15 = (Product of numbers) / (Their H.C.F).
H.C.F of 12 and 15 = 3.
L.C.M = (12 * 15) / 3 = 60.

3. The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275, find the other number.

Product of two numbers = H.C.F * L.C.M.
Let the other number be x.
275 * x = 11 * 7700.
x = (11 * 7700) / 275 = 308.

4. Find the L.C.M of 6, 8, and 12.

L.C.M of 6, 8, and 12 can be found by prime factorization.
6 = 2 * 3, 8 = 2³, 12 = 2² * 3.
L.C.M = 2³ * 3 = 24.

5. Two numbers are in the ratio 4:5. Their L.C.M is 180. Find the numbers.

Let the numbers be 4x and 5x.
Their L.C.M = 4x * 5x / H.C.F of (4x, 5x) = 180.
Since H.C.F of 4 and 5 is 1, L.C.M = 20x = 180.
x = 9.
Numbers are 4x = 36, 5x = 45.

6. The H.C.F of two numbers is 17 and their sum is 136. If one number is 51, find the other number.

Let the other number be x.
x + 51 = 136.
x = 136 - 51 = 85.

7. What is the least number that when divided by 20, 25, 35, and 40 leaves the same remainder 7 in each case?

Least number = L.C.M of (20, 25, 35, 40) - Remainder.
L.C.M of (20, 25, 35, 40) = 1400.
Least number = 1400 - 7 = 1393.

8. Find the greatest number that will divide 43, 91, and 183 so as to leave the same remainder in each case.

Required number = H.C.F of (91 - 43), (183 - 91), and (183 - 43).
= H.C.F of 48, 92, and 140.
= 4.

9. The H.C.F of two numbers is 23 and the other two factors of their L.C.M are 13 and 14. What are the two numbers?

The numbers are 23 * 13 and 23 * 14 = 299 and 322.

10. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, and 12 seconds respectively. In 30 minutes, how many times do they toll together?

L.C.M of 2, 4, 6, 8, 10, 12 = 120 seconds.
In 30 minutes (1800 seconds), they toll together 1800/120 + 1 = 15 + 1 = 16 times.
Subtracting the first toll, they toll together 15 times.

11. Find the least number which when divided by 15, 20, and 25 leaves a remainder of 8 in each case.

Least number = L.C.M of (15, 20, 25) - Remainder.
L.C.M of (15, 20, 25) = 300.
Least number = 300 + 8 = 308.

12. The L.C.M of two numbers is 864 and their H.C.F is 144. If one number is 288, what is the other number?

Product of two numbers = H.C.F * L.C.M.
Let the other number be x.
288 * x = 144 * 864.
x = (144 * 864) / 288 = 432.

13. What is the smallest number that is divisible by each of the numbers 1 to 10?

The smallest number divisible by each of the numbers 1 to 10 is their L.C.M.
L.C.M of 1 to 10 = 2520.

14. A number is divisible by both 6 and 15. What is the smallest such number that is greater than 50?

The smallest number divisible by both 6 and 15 is their L.C.M.
L.C.M of 6 and 15 = 30.
The smallest such number greater than 50 is 60.

15. The L.C.M of three different numbers is 120. Which of the following cannot be their H.C.F?

The H.C.F of numbers cannot be a factor that does not divide the L.C.M. 120 is not divisible by 14, so 14 cannot be their H.C.F.

16. Two numbers are in the ratio 3:4. If their L.C.M is 84, find the numbers.

Let the numbers be 3x and 4x.
Their L.C.M = 3x * 4x / H.C.F of (3x, 4x) = 84.
Since H.C.F of 3 and 4 is 1, L.C.M = 12x = 84.
x = 7.
Numbers are 3x = 21, 4x = 28.

17. The greatest number that will divide 135, 225 and 315 leaving the same remainder in each case is:

Required number = H.C.F of (225 - 135), (315 - 225), and (315 - 135).
= H.C.F of 90, 90, and 180.
= 45.

18. If the H.C.F of two numbers is 12 and their L.C.M is 336, what is the sum of the two numbers?

Let the numbers be a and b.
H.C.F * L.C.M = a * b.
12 * 336 = a * b.
The sum of numbers = a + b.
Possible pairs (a, b) are (48, 84) and (84, 48).
Sum = 48 + 84 = 132.

19. The L.C.M of two numbers is 48. The numbers are in the ratio 2:3. What is the smaller number?

Let the numbers be 2x and 3x.
Their L.C.M = 2x * 3x / H.C.F of (2x, 3x) = 48.
Since H.C.F of 2 and 3 is 1, L.C.M = 6x = 48.
x = 8.
Smaller number = 2x = 16.

20. What is the least number which when divided by 5, 6, 7, and 8 leaves a remainder of 3, but when divided by 9 leaves no remainder?

Least number = L.C.M of (5, 6, 7, 8) + 3.
L.C.M of (5, 6, 7, 8) = 840.
Least number = 840 + 3 = 843.
843 is not divisible by 9. The next multiple of 840 divisible by 9 is 1680.
Required number = 1680 + 3 = 1683.