Let the unit digit be p and ten’s digit be q.
∴ Number = 1000q + 100p + 10q + p
Number = 1010q + 101p = 101(10q + p)

Correct Option: D

Let the unit digit be p and ten’s digit be q.
∴ Number = 1000q + 100p + 10q + p
Number = 1010q + 101p = 101(10q + p)
Clearly, this number is divisible by 101, which is the smallest three-digit prime number.

n²(n²–1) = n² (n + 1) (n – 1)
Now, we put values n = 2, 3..... When n = 2
∴  n²(n² –1) = 4 × 3 × 1 = 12, which is a multiple of 12

Correct Option: B

n²(n²–1) = n² (n + 1) (n – 1)
Now, we put values n = 2, 3..... When n = 2
∴  n²(n² –1) = 4 × 3 × 1 = 12, which is a multiple of 12
When n = 3.
n²(n² –1) = 9 × 4 × 2 = 72,
which is also a multiple of 12. etc.

A number is divisible by 11 if the difference of the sum of digits at odd and even places be either zero or multiple of 11.

Correct Option: D

A number is divisible by 11 if the difference of the sum of digits at odd and even places be either zero or multiple of 11. If the middle digit be 4, then 24442 or 244442 etc are divisible by 11.

A number is divisible by 11, if the difference of sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

Correct Option: D

A number is divisible by 11, if the difference of sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.
Difference

A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits of even places, is either 0 or a number divisible by 11.
∴  (5 + 9 + * + 7) – (4 + 3 + 8) = 0 or multiple of 11

Correct Option: D

A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits of even places, is either 0 or a number divisible by 11.
∴  (5 + 9 + * + 7) – (4 + 3 + 8) = 0 or multiple of 11
⇒  21 + * – 15
∴  * + 6 = a multiple of 11
∴  * = 5