If the students of a class can be grouped exactly into 6 or 8 or 10, then
Required number of students = LCM of 6, 8, 10

Correct Option: B

If the students of a class can be grouped exactly into 6 or 8 or 10, then
Required number of students = LCM of 6, 8, 10 = 120

We find LCM of = 10, 16, 24

∴ LCM = 2² × 2² × 3 × 5
∴ Required number = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

Correct Option: D

We find LCM of = 10, 16, 24

∴ LCM = 2² × 2² × 3 × 5
∴ Required number = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
Required number = 3600

We have , The LCM of 12, 18, 21, 30

∴ LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260

Correct Option: C

We have , The LCM of 12, 18, 21, 30

∴ LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260

LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022

Correct Option: B

LCM of 4, 6, 10, 15 = 60
Least number of 6 digits = 100000
The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020
∴ Required number (N) = 100020 + 2 = 100022
Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5

We know that when a number is divided by a, b or c leaving remainders p, q or 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 = 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3

Correct Option: A

We know that when a number is divided by a, b or c leaving remainders p, q or 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 = 4 – 1 = 3, 5 – 2 = 3, 6 – 3 = 3
∴ The required number = k - t = LCM of ( 4, 5, 6 ) – 3
Hence , The required number = 60 – 3 = 57