LCM and HCF
-  The smallest number, which, when divided by 12 or 10 or 8, leaves remainder 6 in each case, is
- 
                        View Hint View Answer Discuss in Forum 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 = 6Correct Option: CAs 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
-  When a number is divided by 15, 20 or 35, each time the remainder is 8. Then the smallest number is
- 
                        View Hint View Answer Discuss in Forum 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 = 8Correct Option: AAs 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
-  The greatest number of four digits which when divided by 12, 16 and 24 leave remainders 2, 6 and 14 respectively is
- 
                        View Hint View Answer Discuss in Forum 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 = 9984Correct Option: AAs 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
-  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
- 
                        View Hint View Answer Discuss in Forum 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 + 3Correct Option: BThe 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
-  The least number, which when divided by 18, 27 and 36 separately leaves remainders 5,14, and 23 respectively, is
- 
                        View Hint View Answer Discuss in Forum 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 = 108Correct Option: AThe 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
 
	