### Theory:

Similar to the logics on two-digit numbers, three-digit numbers can also be subjected to a few tricks.
Trick $$1$$: What happens to the difference between three-digit numbers and their reverse?
The difference between three digit numbers and their reverse is always a multiple of $$99$$.

Consider the three-digit number '$$abc$$'.

The general form of '$$abc$$' is $$(100 \times a) + (10 \times b) + c$$ or $$100a + 10b + c$$.

The reverse of '$$abc$$' is '$$cba$$'.

The general form of '$$cba$$' is $$(100 \times c) + (10 \times b) + a$$ or $$100c + 10b + a.$$.

Now, let us find the difference between '$$abc$$' and '$$cba$$'.

If $$a > c$$:

$$abc - cba = (100a + 10b + c) - (100c + 10b + a)$$

$$= 100a + 10b + c - 100c - 10b - a$$

$$= 99a - 99c$$

$$= 99(a - c)$$

If $$c > a$$:

$$cba - abc = (100c + 10b + a) - (100a + 10b + c)$$

$$= 100c + 10b + a - 100c - 10b - a$$

$$= 99c - 99a$$

$$= 99(c - a)$$
Example:
Let us impose this logic on $$325$$.

The reverse of $$325$$ is $$523$$.

Hence, to find the difference, we should subtract $$325$$ from $$523$$.

$$523 - 325 = 198$$

$$= 99 \times 2$$

Hence, it is proved that the difference between three-digit numbers and their reverse is a multiple of $$99$$.
Trick $$2$$: What happens when $$3$$ forms of a three-digit number is summed up?
Consider the number '$$abc$$'.

Form $$1$$: $$abc = 100a + 10b + c \longrightarrow (1)$$

To find the other forms of the number, shift the ONES digit to the number's left end.

Therefore, the number '$$abc$$' becomes '$$cab$$'.

The digit '$$c$$' is shifted to the left end of the number.

Form $$2$$: $$cab = 100c + 10a + b \longrightarrow (2)$$

To form the third number, shift '$$b$$' from '$$cab$$' to the left end.

Form $$3$$: $$bca = 100b + 10c + a \longrightarrow (3)$$

On adding $$(1)$$, $$(2)$$, and $$(3)$$, we get:

$$abc + cab + bca = 100c + 10a + b + 100b + 10c + a + 100b + 10c + a$$

$$= 111(a + b + c)$$

$$= 37 \times 3(a + b + c)$$

Therefore, the sum of $$3$$ forms of a three-digit number is always a multiple of $$37$$.
Example:
Let us consider the number $$128$$.

Form $$1$$: $$128$$

Form $$2$$: $$812$$

Form $$3$$: $$281$$

On adding the $$3$$ numbers, we get:

$$128 + 812 + 281 = 1221$$

$$= 37 \times 33$$

Thus, the sum of three forms of a three-digit number is always a multiple of $$37$$.