Number System Divisibility Shortcut Tricks

Divisibility questions are asked in Most of the competitive and Entrance examinations including SSC, IBPS, TET and others. These questions are asked in number system in quantitative aptitude or math section. 

These questions are one of the most difficult and time consuming for candidates who had not solved them before. But the difficulty of these Divisibility questions in Number System can be eased with the help of shortcut tricks which I am going to mention here.
Suppose you are given a Number System-Divisibility question like below one.

Example Question 1: Determine the remainder when 24^81 is divided by 23? (^ means power)
Solution with Shortcut trick:
To solve the question, first divide 24 by 23. The remainder is 1.
Now you have (1)^81 which equals 1. So the remainder is 1. Ans.

Important Note: You have to convert Numerator to 1 or -1 value by dividing it with denominator. We have applied this formula in the above and below questions.

Example Question 2:
Determine the remainder when 2^24 is divided by 7?
Easy Solution:
Given (2^24)/7
Now convert we have to convert the numerator to 1 or -1 in relation to denominator.
So convert (2^24)/7 to [(2^3)^21]/7
So that it becomes (8^21)/7
Now when you divide 8 by 7, the remainder is 1.
So we get 1^21 which equals 1.
So the remainder is 1. Ans.
Let us take another example to better understand this trick.
Example 3. What is the remainder when 51^202 is divided by 7.
Here is Easy shortcut for this.
Divide 51 by 7. The remainder is 2.
So we have (2^202)/7
Which equals [(2^3)^67x (2^1)]/7
Since 2^3 equal 8. Dividing it with 7, we get remainder 1.
So it becomes [(1^67)x2]/7
1^67 equals 1. Multiplying it with 2, we get 2.
So we get 2/7. So the remainder is 2.
Note: When a small numerator is divided by large denominator, the remainder is always that numerator i.e. for 1/7, the remainder is 1.
For 3/7, the remainder is 3. For 5/9, the remainder is 5.

LCM & HCF Divisibility Trick

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