Basics: Divisibility test
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 num1*num2*num3*num4.................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.
Click the button to check the output
Example: Find whether the number 55277838 is divisible by 7 or not.
Solution:
For any modifications/additional topics related to this, please let us know through comments.
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 num1*num2*num3*num4.................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.
Tool to check divisibility test
Enter Divisor and dividendClick the button to check the output
Example: Find whether the number 55277838 is divisible by 7 or not.
Solution:
| Previous State | Input | [ (previous State + Input)*3 ]%7= new state |
|---|---|---|
| 0 | 5 | (0+5)*3%7 = 15%7 ==1 |
| 1 | 5 | (1+5)*3%7 = 18%7 ==4 |
| 4 | 2 | (4+2)*3%7 = 18%7 ==4 |
| 4 | 7 | (4+7)*3%7 = 33%7 ==5 |
| 5 | 7 | (5+7)*3%7 = 36%7==1 |
| 1 | 8 | (1+8)*3%7 = 27%7 ==6 |
| 6 | 3 | (6+3)*3%7 = 27%7 ==6 |
| 6 | 8 | (6+3)*3%7 = 27%7 ==0 => DIVISIBLE |
For any modifications/additional topics related to this, please let us know through comments.
Thanks for reading :)
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