Any number is divisible by 11 when the differences of alternative digits is 0 or multiple of 0, 11 etc. Here,

Correct Option: C

Any number is divisible by 11 when the differences of alternative digits is 0 or multiple of 0, 11 etc. Here,

Therefore , the place of * will come 5.

As we know that If a number is divisible by both 11 and 13, then it will be also divisible by multiple of both 11 and 13 .

Correct Option: C

As we know that If a number is divisible by both 11 and 13, then it will be also divisible by multiple of both 11 and 13 i.e.
divisible by (11 × 13) .

Number = 100p + 10q + r
Sum of digits = p + q + r
Difference = 100p + 10q + r – p – q – r

Correct Option: C

Number = 100p + 10q + r
Sum of digits = p + q + r
Difference = 100p + 10q + r – p – q – r
Difference = 99p + 9q = 9 (11p + q)
Therefore , required answer is 9 .

n³ – n = n (n + 1) (n – 1)
n = 1, n³ – n = 0
n = 2, n³ – n = 2 × 3 = 6

Correct Option: C

n³ – n = n (n + 1) (n – 1)
n = 1, n³ – n = 0
n = 2, n³ – n = 2 × 3 = 6
n = 3, n³ – n = 3 × 4 × 2 = 24
n = 4, n³ – n = 4 × 5 × 3 = 60
⇒ 60 ÷ 6 = 10
Hence , required answer is 6.

n³ – n = n (n² – 1)
n³ – n = n (n + 1) (n – 1)
For n = 2, n³ – n

Correct Option: B

n³ – n = n (n² – 1)
n³ – n = n (n + 1) (n – 1)
For n = 2, n³ – n
⇒ 2³ – 2 = 8 - 2 = 6
Hence , ( n³ – n ) is divisible by 6.