LCM of 4, 6, 8, 9

∴ LCM = 2 × 2 × 3 × 2 × 3 = 72

Correct Option: B

LCM of 4, 6, 8, 9

∴ LCM = 2 × 2 × 3 × 2 × 3 = 72
∴ Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13.

As we know that when a number is divided by a, b leaving remainders p, q respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a and b .
Here, t = 12 – 5 = 7, 16 – 9 = 7

Correct Option: B

As we know that when a number is divided by a, b leaving remainders p, q respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a and b .
Here, t = 12 – 5 = 7, 16 – 9 = 7
∴ Required number = k - t = (L.C.M. of 12 and 16) – 7
Required number = 48 – 7 = 41

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 = Divisor – remainder = 1
t = 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1

Correct Option: C

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 = Divisor – remainder = 1
t = 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1
∴ Required number = k - t = (L.C.M. of 10, 9, 8) – 1
Hence , Required number = 360 – 1 = 359

We find LCM of 5, 6 and 8
5 = 5
6 = 3 × 2
8 = 2³
LCM of 5, 6 and 8 = 2³ × 3 × 5 = 8 × 15 = 120
Required number = 120K + 3
when K = 2,

Correct Option: D

We find LCM of 5, 6 and 8
5 = 5
6 = 3 × 2
8 = 2³
LCM of 5, 6 and 8 = 2³ × 3 × 5 = 8 × 15 = 120
Required number = 120K + 3
when K = 2,
∴ Required number = 120 × 2 + 3 = 243
It is completely divisible by 9

LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
∴ Required number = 1936 – 1897

Correct Option: C

LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
∴ Required number = 1936 – 1897 = 39