## LCM and HCF

#### LCM and HCF

1. The smallest number, which, when divided by 12 or 10 or 8, leaves remainder 6 in each case, is

1. As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
The smallest number divisible by 12 or 10 or 8
= LCM of 12, 10 and 8 = 120
remainder = 6

##### Correct Option: C

As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
The smallest number divisible by 12 or 10 or 8
= LCM of 12, 10 and 8 = 120
remainder = 6
∴ Required number = 120 + 6 = 126

1. When a number is divided by 15, 20 or 35, each time the remainder is 8. Then the smallest number is

1. As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
LCM of 15, 20 and 35 ( k ) = 420
Here , remainder = 8

##### Correct Option: A

As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
LCM of 15, 20 and 35 ( k ) = 420
Here , remainder = 8
∴ Required least number = 420 + 8 = 428

1. The greatest number of four digits which when divided by 12, 16 and 24 leave remainders 2, 6 and 14 respectively is

1. As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
Here, t = 12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10
Now, LCM of 12, 16 and 24 = 48
∴ The greatest 4–digit number exactly divisible by 48 = 9984

##### Correct Option: A

As we know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
Here, t = 12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10
Now, LCM of 12, 16 and 24 = 48
∴ The greatest 4–digit number exactly divisible by 48 = 9984
∴ Required number = 9984 – 10 = 9974

1. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder is

1. The LCM of 5, 6, 7 and 8 = 840
∴ Required number = 840k + 3 , which is exactly divisible by 9 for some value of k.
Now, 840k + 3 = 93 × 9k + (3k + 3)
When k = 2 , 3k + 3 = 9, which is divisible by 9.
∴ Required number = 840k + 3

##### Correct Option: B

The LCM of 5, 6, 7 and 8 = 840
∴ Required number = 840k + 3 , which is exactly divisible by 9 for some value of k.
Now, 840k + 3 = 93 × 9k + (3k + 3)
When k = 2 , 3k + 3 = 9, which is divisible by 9.
∴ Required number = 840k + 3 = 840 × 2 + 3 = 1683

1. The least number, which when divided by 18, 27 and 36 separately leaves remainders 5,14, and 23 respectively, is

1. The difference between the divisor and the corresponding remainder is same in each case
ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13
∴ Required number = (LCM of 18, 27, and 36 ) – 13
LCM of 18, 27, and 36 = 108

##### Correct Option: A

The difference between the divisor and the corresponding remainder is same in each case
ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13
∴ Required number = (LCM of 18, 27, and 36 ) – 13
LCM of 18, 27, and 36 = 108
Hence , Required number = 108 – 13 = 95