Euler's Remainder Theorem

Let’s understand this theorem with an example:

Q.5) – Rahul has certain number of cricket balls with him. If he divides them into 4 equal groups, 2 are left over. If he divides them into 7 equal groups, 6 are left over. If he divides them into 9 equal groups, 7 are left over. What is the smallest number of cricket balls could Rahul have?

Let N be the number of cricket balls.

N = 2(mod4) ————–> equation 1

N = 6(mod7) ————–> equation 2 &

N = 7(mod9) ————–> equation 3.

From N=2(mod4) we get, N=4a+2

Substituting this in equation 2, we get the following equation:

4a + 2 = 6(mod7)

Therefore, 4a = 4(mod7)

Hence, 2 x 4a = 2 x 4(mod7)

This gives us a = 1(mod7)

Hence a = 7b+1.

Plugging this back to N=4a+2, we get….

N = 28b + 6

Substituting this to equation 2;

28b + 6 = 7(mod9)

28b = 1(mod9)

Therefore, b=1(mod9)

Hence b = 9c + 1.

Substituting this back to equation N=28b+6;

N = 28(9c+1) + 6

N = 252c + 34

The smallest positive value of N is obtained by setting c=0.

It gives us N = 34

TB – All prime numbers greater than 3 can be expressed as 6K+1 or 6K-1, this is another important result. You would be using this result a lot when it comes to number system problems...

Q) If n is a positive integer, what is the remainder when [7(8n+3) + 2] is divided by 5?

This problem can be easily solved with concept of cyclicity.

You should realize that unit digit of all numbers raised to powers start repeating itself.

For example, the cyclicity of 7 is 4. This means that the unit digits repeat after on every 4th power. See for yourself below:

71 = 7 75 =7 79=7

42 = 9 76= 9 710=9

73 = 3 77 =3 711=3

74 = 1 78 =1

In this question, consider n=1.

You are looking at unit digit of 7^11 which will be 3 from concept of cyclicity.

Add two to it.

Therefore, last digit will be 5 for [7(8n+3) + 2]. Hence it’s completely divisible by 5.

Remainder can never be negative; its minimum value can only be 0.

Consider an example of -30 / 7. Here, remainder is 5.

It would not be (-28 – 2 / 7), but [(-35+5)/7]

When you divide, you will get remainder of -2. Since remainder can never be negative, we subtract it from quotient, here 7 – 2 = 5.

Negative remainder is useful when you are trying to solve a problem with higher power.

Consider an example…….

Q) Find the remainder for 7^7^7 is divided 32

32 can be factored into 8 and 4,

Therefore, we will divide the question into two parts:

Remainder [7^7^7/8] x Remainder [7^7^7/4]

which gives: Remainder [(-1) ^7^7/8] x Remainder [3^7^7/4] [(-1) raised to odd power is -1 and raised to even power is 1]

Remainder [(-1)/8] x Remainder [(-1)/4]…. which gives

———-> 7 x 3 = 21

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