Many problems in Aptitude solving can be done in seconds if you can apply these rules for divisibility test.
Divisible by: Rule
2 last digit is even(0,2,4,6,8)
3 - sum of all the digits is divisible by 3
4 last 2 digits should be divisible by 4
5 the last digit is either 0 or 5
6 should be divisible by both 2 and 3
8 last 3 digits should be divisible by 8
9 sum of all digits of the number should be divisible by 9
10 last digit should be 0
11 difference between sum of odd digits and sum of even digits should be zero
In the case of 7, there is a small procedure:
When we divide 10,100,1000............... by 7, the remainder is 3. so there is a small formula for divisibility by 7.
New state resembles the remainder for the number by 7.
[ (previous State + Input)3 ]%7 = new state.
where the previous state for the first time is ZERO and Input one by one digit.
In general if num1num2num3num4.................num_n=p,
divisibility test for p is it is divisible by all num1,num2,num3,num4...............num_n.
for example divisibility by 122, test number should be divisible by both 2 and 61
with this type of minimization, we can simplify it more.