Alright, here are the solutions that I got.

1. It is given that the two digits of the required number are prime numbers i.e. 2, 3, 5 or 7. Note that 1 is neither prime nor composite. Also, the third digit is the multiplication of the first two digits. Thus, hundreds digit and units digit must be either 2 or 3 i.e. 2_2, 2_3, 3_2 or 3_3 which means that there are four possible numbers - 242, 263, 362 and 393.

Now, it is also given that - the difference between its reverse and it is 99. So 263 and 362 satisfy this condition. Hence, the sum of the three digits is 11 in each case.

2. It is given that the man has 40 litres container of milk. Also, he will drink 1 litre on the first day and refill the container with water, will drink 2 litres on the second day and refill the container, will drink 3 litres on the third day and refill the container, and so on till 40th day. Thus at the end of 40 days, he must have drunk (1 + 2 + 3 + 4 + ..... +38 + 39 + 40) = 820 litres of liquid.

Out of those 820 litres, 40 litres is the milk which he had initially. Hence, he must have drunk 780 litres of water.

3. As there are 11 balls along one side, it means that there are 11 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55 and 66 balls in the remaining layers.

Hence, the total number of balls is = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286 balls

4. Let's start with JKL = 9 * LQ. Note that L appear on both the side. Also, after multiplying LQ by 9 the answer should have L at the unit's place. The possible values of LQ are 19, 28, 37, 46, 55, 64, 73, 82 and 91; out of which only 64, 73 and 82 satisfies the condition. (As all alphabets should represent different digits)

Now, consider PQR = 6 * LQ. Out of three short-listed values, only 73 satisfies the equation.

Also, ZYX = 3 * LQ is satisfied by 73.

Hence, Z=2, Y=1, X=9, P=4, Q=3, R=8, J=6, K=5, L=7

219/3 = 438/6 = 657/9 = 73

5. The old man temporarily added his dog to the 17, making a total of 18 dogs.

First son got 1/2 of it = 9

Second son got 1/3 of it = 6

Third son got 1/9 of it = 2 for a total of 17.

He then steals his dog back and goes away......