Solution:

The integer with the special property is -1.

To test the property, let us subtract any digit from it, let's say 5.

-1 - 5 = -6

The succeeding digit of 5 is 6.

When (-6) is divided by 6, it is completely divided.

For all other numbers as well, this stands true.

Solution:

The required number is 499449.

The maximum sum of a number under 5 lakh will be of the number 499999 i.e. 49.
Thus the required number must be close to 499999.
The root of 499999 ~ 707.106
By calculating the square of 707, we get 499849 which is the answer.

Solution:

2418

Explanation:
The rule that is being followed here is

P ~ Q ~ R ~ S = [first digit of (P * S)] [last digit of (Q * R)] [first digit of (Q * R)] [last digit of (P * S)]

Using the same rule in the incomplete equation,
7 ~ 2 ~ 7 ~ 4 = [first digit of (7 * 4)] [last digit of (2 * 7)] [first digit of (2 * 7)] [last digit of (7 * 4)]
= 2418

Thus 2418 is the missing number.

Solution:

Girl

Explanation:
In the House P:
There can be three possibilities.
Older Younger
Boy Girl
Girl Boy
Boy Boy
Since a boy is mentioned, we can't have a Girl-Girl option. Now all these options are equally possible and thus we will take them all into account. Two of the outcome have girls as an option and thus the probability here will be 2/3.

In the House Q:
We know that the elder child is a boy and thus there can be only two possibilities:
Older Younger
Boy Girl
Boy Boy

Here, the probability of having a girl child will be 1/2.

Thus you must choose the house P as we have better probability of having a girl in that house.

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.

Solution:

$6 for red sponges = 1 sponges
$87 for yellow sponges = 29 sponges
$7 for blue sponges = 70 sponges

Explanation:
a = red sponges
b = yellow sponges
c = blue sponges

Now using a simple expression:
A + b + c = 100
6a + 3b + 0.1c = 100

Now if we multiply the first equation by 6 and then 3 and then subtracting the two equations, we will be able to get two more equations:

6a + 6b + 6c = 600
3a + 3b + 3c = 300

3b = 500 - 5.9c
3a = 2.9c - 200

Now if you are wondering how will we solve three variables using just two equations, we have to tell you that a and b are non-negative integers. Thus, if 3b >/= 0, then 500 - 5.9c >/= 0.
Using this, c Also, if 3a >/= 0, then 2.9c - 200 > 0. This implies that c >/= 68.97.

But buying the blue sponges is the only way to spend in fraction thus the number of blue sponges that we buy must cost an even dollar amount. Now if you consider the expressions, there can be only two numbers we can buy between 68.97 and 84.75 that satisfies the given condition i.e. 70 and 80.

Let's say that we replace c by 80 in the last two expressions that we have formed above, a will equal to 10.67 and b will equal to 9.33. However, the values of a and b must be an integer and not fraction. Therefore, c can't be 80.

Let us now replace c with 70. After solving the expressions, we will obtain a as 1 and b as 29.

Thus here are the amounts that must be used to buy:
$6 for red sponges = 1 sponges
$87 for yellow sponges = 29 sponges
$7 for blue sponges = 70 sponges

Solution:

10 = 9.9999999... .
The proof is correct and the fact is for real. There is no error in it.

Solution:

192

Explanation:
I started with the number 192.
Adding the reversed 291 to 192 gave me 483.
Similarly,
483 + 384 = 867
867 + 768 = 1635
1635 + 5361 = 6996

Solution:

3816547290

Solution:

4

Explanation:
Now if you go on calculating 2^300, you will lose so much time and it might not even be feasible to carry out so long of calculations. Thus we will make use of a trick here. We will calculate the remainder of each power of 2 till we come across a pattern.

2^1 divided by 4 leaves the remainder 2.
2^2 divided by 4 leaves the remainder 0.
2^3 divided by 4 leaves the remainder 0.
2^4 divided by 4 leaves the remainder 0.

You can see that all the following powers of two will be divisible by 4. Therefore the remainder when 2^300 is divided by 4 will be 0 only.