HCF and LCM - Practice Questions with Tricks

HCF and LCM

Example 1: The LCM of two numbers is 864 and their HCF is 144. If one of the number is 288, the other number is?

Solution: Product of two numbers = HCF x LCM
        ⇒ 288 x second number = 864 x 144
        ⇒ Second number = (864 x 144)/288
        ⇒ 432

Example 2: The HCF and LCM of two 2- digit numbers are 16 and 480 respectively. The numbers are?

Solution: We know HCF is a factor of number itself. So two numbers can be expressed as 16x and 16y, where x and y are prime to each other.
Now, Product of numbers = HCF x LCM
  ⇒ 16x x 16y = 16 x 480
  ⇒ xy = (16 x 480)/(16 x 16) = 30
Possible pairs of x and y will be = (1, 30); (2,15); (3, 10); (5, 6)

According to question, numbers are of two digits. Therefore, only possible pair is (5, 6).
  ⇒ Numbers are = 16x = 16 x 5 = 80
                                & 16y = 16 x 6 = 96

Example 3: The product of two numbers is 2028 and their HCF is 13. The number of such pairs is?

Solution: We know HCF is a factor of number itself. So two number can be expressed as 13x and 13y, where x and y are prime to each other.
Now Product of two numbers = 2028
  ⇒ 13x x 13y = 2028
  ⇒ xy = 2028/169 = 12
Possible pairs of x and y will be = (1, 12); (2,6); (3, 4)
But 2 & 6 are not co-prime numbers, so the required number of pairs are 2 [(1, 12) & (3,4)]

Example 4: Find the least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case?

Solution: We know that, on dividing a number by a, b and c if we get 'k' as remainder, then the number will be = n x LCM of [a, b, c] + k
  ⇒ LCM of (4, 6, 8, 12, 16) = 48
  ⇒ Number will be = 48 + 2 = 50

Example 5: Find the greatest number of five digits which when divided by 3, 5, 8 and 12 have 2 as remainder?

Solution: We know that, on dividing a number by a, b and c if we get 'k' as remainder, then the number will be = n x LCM of [a, b, c] + k
LCM of (3, 5, 8, 12) = 120

Greatest number of five digits = 99999
"If after dividing any number 'A' by LCM of given numbers (x, y, z) we get remainder 'R', then those numbers will also give same remainder 'R' after dividing the given number 'A' separately."

After dividing 99999 by 120, we get 39 as remainder.
⇒ 99999-39 = 99960 is the greatest five digit number which will be divisible by 120 and given divisors.

In order to get 2 as remainder simply add 2 to the given number.
  ⇒ Required number = 99960 + 2 = 99962
[99962 = 833 x 120 + 2 = n x LCM of (3, 5, 8, 12) + 2]

Example 6: A, B, C start running at the same time and at the same point in the same direction in a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds. After what time will they meet again at the starting point?

Solution: We know that LCM of two or more numbers is the least number which is exactly divisible by each of them.
Here the time at which all three will meet again is the Least time which is LCM of the time taken by A, B and C in completing a round.

  ⇒ LCM of (252, 308, 198) = 2772 seconds = 46 minutes 12 seconds

Example 7: Find the largest number of four digits such that on dividing by 15, 18, 21 and 24 the remainders are 11, 14, 17 and 20 respectively?

Solution: Here (15 - 4 = 11); (18 - 4 = 14); (21 - 4 = 17); (24 - 4 = 20)
We know that, on dividing a number by a, b & c if we get (a - k), (b - k), (c - k) as remainder, then that number will be = n x LCM of [a, b, c] – k
LCM of (15, 18, 21, 24) = 2520

Largest number of 4 digits = 9999
"If after dividing any number 'A' by LCM of given numbers (x, y, z) we get any remainder 'R', then those numbers will also give same remainder 'R' after dividing the given number 'A' separately."

After dividing 9999 by 2520 we get 2439 as remainder.
  ⇒ 9999-2439 = 7560 will be completely divisible by 2520 and given numbers.

Required number to get respective remainders = 7560 - 4 = 7556
[7556 = 3 x 2520 – 4 = n x LCM of (15, 18, 21, 24) – k]

Example 8: Find the least perfect square, which is divisible by each of 21, 36 and 66?

Solution: 21 = 3 x 7
                  36 = 2 x 2 x 3 x 3
                  66 = 2 x 3 x 11
LCM of 21, 36 and 66 = 2 x 2 x 3 x 3 x 7 x 11

To make it a least perfect square, multiply it by 7 x 11.
Required number = 2 x 2 x 3 x 3 x 7 x 7 x 11 x 11 = 213444

Example 9: Find the least number which when divided by 3, 5, 6, 8, 10 and 12 leaves in each case a remainder of 2 but when divided by 13 leaves no remainder?

Solution: We know that, on dividing a number by a, b and c if we get k as remainder, then the number will be = n x LCM of [a, b, c] + k
Required number = n x LCM of (3, 5, 6, 8, 10, 12) + 2 = 120n + 2

To find the number which will give zero remainder when divided by 13, we need to find the value of n.

The given number (120n + 2) can be written as = 13 x 9n + 3n + 2
Clearly (3n + 2) should be divisible by 13 to make the given number divisible by 13.
For n = 8, (3n + 2) = 26 which is divisible by 13.
Therefore required number = 120n + 2 = 120 x 8 + 2 = 722

Example 10: Find the smallest number, which when increased by 5 is divisible by each of 24, 32, 36 and 54?

Solution: If a number after adding k is divisible by a, b and c then that number will be = n x LCM of [a, b, c] – k.
Required number = LCM of (24, 32, 36, 54) – 5 = 864 - 5 = 859

Example 11: What is the greatest number that will divide 307 and 330 leaving remainders 3 and 7 respectively?

Solution: We know that, Greatest number which divides the number x, y and z leaving remainder a, b and c respectively = HCF of (x-a), (y-b), (z-c)
Number = HCF of (307 - 3) & (330 - 7)
                = HCF of 304 & 323 = 19

Example 12: What is the largest number which divides 25, 73 and 97 to leave the same remainder in each case?

Solution: Greatest number which will divide x, y and z leaving the same remainder in each case is = HCF of (x-y), (y-z), (z-x).
Required number = HCF of |25-73|, |73-97| & |97-25|
                                 = HCF of 48, 24 & 24 = 24

Example 13: What is the least number of square tiles required to pave the floor of a room 15m 17 cm long and 9m 2cm broad?

Solution: Length of the floor = 15m 17cm = 1517 cm
Breadth of the floor = 9m 2cm = 902 cm
Area of the floor = 1517 x 902 cm2

The number of square tiles will be least when the size of each tile is maximum.
Size of each tile = HCF of 1517 and 902 = 41cm
Required number of tiles = (1517x902)/412 = 814

Example 14: Two numbers are in the ratio 3:4. If their HCF is 4, then their LCM will be?

Solution: If HCF of two numbers is x and numbers are ax and bx, then their LCM will be = abx
Here Numbers are 3x and 4x. where x = HCF which is 4.
LCM will be = 3 x 4x =12 x 4 = 48

Example 15: The sum of a pair of positive integer is 336 and their HCF is 21. The number of such pairs will be?

Solution: Let the other factors of number be x and y. Then Numbers will be 21x and 21y.
  ⇒ 21x + 21y = 336
  ⇒ x + y = 336/21 = 16
Possible pairs are = (1, 15); (5, 11); (7, 9); (9,13)