First 3–digit number divisible by 6 = 102
Last such 3-digit number =996
∴ 996 = 102 + (n –1) 6
⇒  (n – 1)6 = 996 – 102 = 894

Correct Option: B

First 3–digit number divisible by 6 = 102
Last such 3-digit number =996
∴ 996 = 102 + (n –1) 6
⇒  (n – 1)6 = 996 – 102 = 894

A number is divisible by 9 and 6 both, if it is divisible by LCM of 9 and 6 i.e., 18.

Correct Option: B

A number is divisible by 9 and 6 both, if it is divisible by LCM of 9 and 6 i.e., 18. Hence, the numbers are 108, 126, 144, 162, 180, 198.
Thus , the total number of integers between 100 and 200 are 6.

A number is divisible by 9 if the sum of its digits is divisible by 9.
Here, 6 + 7 + 0 + 9 = 22

Correct Option: A

A number is divisible by 9 if the sum of its digits is divisible by 9.
Here, 6 + 7 + 0 + 9 = 22
Now, 22 + 5 = 27, which is divisible by 9. Hence 5 must be added to 6709.

A number is divisible by 9, if sum of its digits is divisible by 9.
Let the number be p.
⇒  5 + 4 + 3 + 2 + p + 7

Correct Option: C

A number is divisible by 9, if sum of its digits is divisible by 9.
Let the number be p.
⇒  5 + 4 + 3 + 2 + p + 7
⇒ 21 + p
⇒ 21 + 6 = 27, which is divisible by 9.
∴  p = 6
Therefore , the digit in place of * is 6 .

As we know that ,
xⁿ – aⁿ is exactly divisible by (x – a) if n is odd.

Correct Option: D

As we know that ,
xⁿ – aⁿ is exactly divisible by (x – a) if n is odd.
∴ (49)¹⁵ – (1 )¹⁵ is exactly divisible by 49 – 1 = 48, that is a multiple of 8.
Hence required answer is 8.