Let the 3 digit number be abc. On rearranging the digits I get, abc, bac, cba, cab, acb, bca. Hence I will get 6  three digit numbers.

From 1 to 4 there are 2 numbers divisible by 2, which are 2 and 4. From 1 to 10, there are 5 numbers, which are 2, 4, 6, 8, 10. Hence, the number of numbers divisible by 2 is half the number up to which we count. Half of 100 is 50. Hence 50 is the number of numbers divisible by 2 from 1 to 100.

Let's take a two digit number 'ab'. Its general form is 10 × a + 1 × b. When we  reverse the digits,  the new number will be of the form  10 × b + 1 × a.
If we add both the numbers we get
(10 × a + 1 × b) + (10 × b + 1 × a)
=10a + b + 10b + a
= 11(a+b)

Hence when divided by 11 it will not leave any remainder.

On dividing 1591 by 1000, the remainder is the last 3 digits. Hence it is 591.