In this article, you will learn a quick approach to tackle remainder based questions in Number system. Let's start with an example:

Given 19^{2200002}/23, find the remainder = ?

The apparent complexity of this question can send a chill down the spine of students. However, one concept that makes such questions very simple and easy, is the concept of Euler number.

The Euler number of a number x means the number of natural numbers which are less than x and are co-prime to x. E.g. the Euler number of 6 will be 2 as the natural numbers 1 & 5 are the only two numbers which are less than 6 and are also co-prime to 6.

Mathematically, the Euler number of a number z denoted by the symbol E(z) is calculated as explained below. where P, Q and R are the **different prime factors** of z.

**Euler's Theorem Examples:**

So, the Euler number of 20 will be Hence, there are 8 numbers less than 20, which are co-prime to it. **Cross check:** Numbers co-prime to 20 are 1, 3, 7, 9, 11, 13, 17 and 19, 8 in number.**Application of Euler's Theorem:****Take this test now to assess your understanding of Euler number.****Example 2: ** What is the remainder when 13^{18} is divided by 19?**Solution:** If y^{E (z)} is divided by z, the remainder will always be 1; if y, z are co-prime In this case the Euler number of 19 is 18**Example 3:** What is the remainder when 13^{32} is divided by 15?**Solution:** Now in this case the Euler number of 15 is 8 Now the Numerator can also be written as 13^{8×4}. Thus the remainder in this case will be 1.**Example 4:** Now, let us solve the question given at the beginning of the article using the concept of Euler Number: **Solution: **The Euler Number of the divisor i.e. 23 is 22, where 19 and 23 are co-prime.**Example 5:** How many numbers from 1 to 300 can neither be divisible by 2, nor by 3 nor by 5?**Solution:** The Euler number of 300 is 80, which is the answer.**How ready are you now in Number System? Take this test and get the answer.****Watch this Achievers video on Euler's theorem to grasp all tips & tricks to ace the exam.**###### Key Learning:

**For any query on the topic, feel free to post in the comment section below.**

- This concept has a wonderful application in answering remainder questions.
**When y**Where, E(z) is Euler number of z and y and z are co-prime to each other.^{E (z)}is divided by z, the remainder will always be 1**When y**That is if the power is any multiple of the Euler number of the divisor, even in that case the remainder will be 1.^{E (z)}.k is divided by z, where k is an integer, remainder will always be 1

(The** Euler number of a prime number is always 1 less than the number).** As 13 and 19 are co-prime to each other, **the remainder will be 1.**

What is the remainder of 19^{2200002}/23?

- Hence, the remainder will be 1 for any power which is of the form of 220000.
- The given power is 2200002. Dividing that power by 22, the remaining power will be 2.
- Your job remains to find the remainder of 19
^{2}/23. As you know the square of 19, just divide 361 by 23 and get the remainder as 16.

- Euler's theorem is the most effective tool to solve remainder questions.
- As seen in Example 5, Euler's theorem can also be used to solve questions which, if solved by Venn diagram, can prove to be lengthy.