99 = 1 * 3 * 3 * 11; 101 = 1 * 101;

176 = 1 * 2 * 2 * 2 * 2 * 11; 182 = 1 * 2 * 7 * 13

So, divisors of 99 are 1, 3, 9, 11, 33 and 99;

divisors of 101 are 1 and 101;

divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88 and 176;

divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

1095)1168(1

           1095

        -----------

            73)1095(15

                  1095

               ----------

                      0

So, H.C.F of 1095 and 1168 = 73.

1095/1168 = (1095/73) / (1168/73) = 15/16.

4 * 27 * 3125 = 22 * 33 * 55; 

8 * 9 * 25 * 7 = 23 * 32 * 52 * 7; 

16 * 81 * 5 * 11 * 49 = 24 * 34 * 5 * 72 * 11 

H.C.F = 22 * 32 * 5 = 180.

Explanation:

36 = 22 * 32;   84 = 22 * 3 * 7 

H.C.F = 22 * 3 = 12

2  24 - 36 - 40

--------------------

2  12 - 18 - 20

--------------------

2    6  -  9 - 10

-------------------

3    3  -  9  -   5

--------------------

       1  -  3   -  5

L.C.M = 2 * 2 * 2 * 3 * 3 * 5 = 360. 

Given numbers are 1.08, 0.36 and 0.90.

H.C.F of 108, 36 and 90 is 18.

H.C.F of a given numbers = 0.18

3240 = 23 * 34 * 5; 3600 = 24 * 32 * 52 

H.C.F = 36 = 22 * 32 

Since H.C.F is the product of lowest powers of common factors, so the third number must have (22 * 32 ) as its factor. 

Since L.C.M is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors. 

Third number = 22 * 35 * 72

Let the required numbers be x, 2x and 3x. Then, their H.C.F = x. So, x = 12.

The numbers are 12, 24, 36.

Let the numbers be 3x and 4x.

Then their H.C.F = x. So, x = 4.

So, the numbers are 12 and 16.

L.C.M of 12 and 16 = 48.

Let the required numbers be 33a and 33b. 

Then, 33a + 33b = 528 => a + b = 16.

Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9).

Required numbers are (33 * 1, 33 * 15), (33 * 3, 33 * 13), (33 * 5, 33 * 11), (33 * 7, 33 * 9).

The number of such pairs is 4.