LCM and HCF
 The least perfect square, which is divisible by each of 21, 36 and 66 is

View Hint View Answer Discuss in Forum
LCM of 21, 36 and 66
∴ LCM = 3 × 2 × 7 × 6 × 11
⇒ LCM = 3 × 3 × 2 × 2 × 7 × 11
∴ Required number = 3^{2} × 2^{2} × 7^{2} × 11^{2}Correct Option: C
LCM of 21, 36 and 66
∴ LCM = 3 × 2 × 7 × 6 × 11
⇒ LCM = 3 × 3 × 2 × 2 × 7 × 11
∴ Required number = 3^{2} × 2^{2} × 7^{2} × 11^{2}
Required number = 213444
 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.

View Hint View Answer Discuss in Forum
As per the given details in question , we have
15 = 3 × 5
18 = 3^{2} × 2
21 = 3 × 7
24 = 2^{3} × 3
LCM of 15 , 18 , 21 and 24 = 8 × 9 × 5 × 7 = 2520
The largest number of four digits = 9999Correct Option: B
As per the given details in question , we have
15 = 3 × 5
18 = 3^{2} × 2
21 = 3 × 7
24 = 2^{3} × 3
LCM of 15 , 18 , 21 and 24 = 8 × 9 × 5 × 7 = 2520
The largest number of four digits = 9999
Required number = 9999 – 2439 – 4 = 7556
(Because 15 – 11 = 4
18 – 14 = 4
21 – 17 = 4
24 – 20 = 4)
 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 ?

View Hint View Answer Discuss in Forum
Required time = LCM of 252, 308 and 198 seconds
∴ LCM = 2 × 2 × 7 × 9 × 11Correct Option: D
Required time = LCM of 252, 308 and 198 seconds
∴ LCM = 2 × 2 × 7 × 9 × 11
Required time = 2772 seconds
Required time = 46 minutes 12 seconds
 The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is:

View Hint View Answer Discuss in Forum
According to question ,
LCM of 4, 5, 6, 7 and 8
LCM of 4, 5, 6, 7 and 8 = 2 × 2 × 2 × 3 × 5 × 7 = 840.
Let required number be 840K + 2 which is multiple of 13.
Least value of K for which ( 840K + 2 ) is divisible by 13 is K = 3Correct Option: C
According to question ,
LCM of 4, 5, 6, 7 and 8
LCM of 4, 5, 6, 7 and 8 = 2 × 2 × 2 × 3 × 5 × 7 = 840.
Let required number be 840K + 2 which is multiple of 13.
Least value of K for which ( 840K + 2 ) is divisible by 13 is K = 3
∴ Required number = 840 × 3 + 2
Required number = 2520 + 2 = 2522
 Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder :

View Hint View Answer Discuss in Forum
Here , Remainder ( r ) = 2
As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
The greatest number of five digits is 99999.
LCM of 3, 5, 8 and 12
∴ LCM = 2 × 2 × 3 × 5 × 2 = 120
After dividing 99999 by 120, we get 39 as remainder
99999 – 39 = 99960 = ( 833 × 120 )Correct Option: D
Here , Remainder ( r ) = 2
As we know that When a number is divided by a, b , c or d leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b , c and d.
The greatest number of five digits is 99999.
LCM of 3, 5, 8 and 12
∴ LCM = 2 × 2 × 3 × 5 × 2 = 120
After dividing 99999 by 120, we get 39 as remainder
99999 – 39 = 99960 = ( 833 × 120 )
99960 is the greatest five digit number divisible by the given divisors.
In order to get 2 as remainder in each case we will simply add 2 to 99960.
∴ Greatest number = 99960 + 2 = 99962