Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.

    x = 5k + 3

    x² = (5k + 3)²

   = (25k² + 30k + 9)

   = 5(5k² + 6k + 1) + 4

On dividing x² by 5, we get 4 as remainder.

3-digit number divisible by 6 are: 102, 108, 114,... , 996

This is an A.P. in which a = 102, d = 6 and l = 996

Let the number of terms be n. Then t_n = 996.

a + (n - 1)d = 996

102 + (n - 1) x 6 = 996

6 x (n - 1) = 894

(n - 1) = 149

n = 150

Number of terms = 150.

 4 a 3  |
 9 8 4  }  ==> a + 8 = b  ==>  b - a = 8  
13 b 7  |

Also, 13 b7 is divisible by 11      (7 + 3) - (b + 1) = (9 - b)

  (9 - b) = 0

  b = 9

(b = 9 and a = 1)     (a + b) = 10.

Required numbers are 24, 30, 36, 42, ..., 96

This is an A.P. in which a = 24, d = 6 and l = 96

Let the number of terms in it be n.

Then t_n = 96    a + (n - 1)d = 96

24 + (n - 1) x 6 = 96

(n - 1) x 6 = 72

(n - 1) = 12

n = 13

Required number of numbers = 13.

Marking (/) those which are are divisible by 3 by not by 9 and the others by (X), by taking the sum of digits, we get:s

2133 9 (X)

2343 12 (/)

3474 18 (X)

4131 9 (X)

5286 21 (/)

5340 12 (/)

6336 18 (X)

7347 21 (/)

8115 15 (/)

9276 24 (/)

Required number of numbers = 6.