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Concept of Successive Division

Suppose N is any number which is divided successively by 3 and 5, which means – first, we divide N by 3 and then the quotient obtained is divided by 5.
E.g., Let us see the case when 50 is divided by 5 and 3 successively.
50 divided by 5 gives 10 as the quotient. Now, we will divide 10 by 3. It gives finally a quotient of 3 and remainder of 1.
When a number N is divided successively by 3 and 5, remainder obtained are 1 and 2 respectively. What is the remainder when N is divided by 15?
It can be seen that we are required to calculate it from back-end.
Family of numbers which when divided by 5 gives remainder 2 = 5 S + 2
So, N = 3(5 S + 2) + 1 = 15 S + 7
Now, when N is divided by 15, remainder = 7

Fermat’s remainder therorem

Let P be a prime number and N be a number not divisible by P. Then the remainder obtained when AP-1 is divided by P is 1.
What is the remainder when 2100 is divided by 101?
Since it satisfies the Fermat’s theorem format, remainder = 1

Euler’s theorem of remainder

Let f(n) be the number of integers less than n and co-prime with n, then the remainder obtained when mf(n) is divided by n is 1, where m and n are co-prime to each other.
As we can see:
f(2) = 1, f(3) = 2, f(4) = 2, f(5) = 4
So, the remainder obtained when 8f(5) = 4 is divided by 5 is 1. It can also be seen that the remainder obtained when any power of 8 divisible by 4 such as 84 or 88 or 812 and so on will give the same remainder when divided by 5.
Similarly, the remainder obtained when 12f(5) = 4 is divided by 5 is 1; or, the remainder obtained when 8f(7) = 6 is divided by 7 is 1.


  1. Description: 5691.png will always give 1 as the remainder.
    (For all natural values of A and N)
What is the remainder when 9100 is divided by 8?
For A = 8, it satisfies the above condition. 
So, remainder = 1
Alternatively, we can apply either cyclicity or theorem method to find the remainder. (Do this yourself).
  1. Description: 5685.png When N is even, remainder is 1 and when N is odd, remainder is A itself.
What is the remainder when 210 is divided by 3?
Since here N is even, so remainder = 1
  1. i. (an + bn) is divisible by (a + b), if n is odd.
    Extension of the above formula – (an + bn + cn) is divisible by (a + b + c), if n is odd and a, b, c, are in arithmetic progression.
    Similarly, the above situation can be extended for any number of terms.
    1. (an  bn) is divisible by (a + b), if n is even.
    2. (an  bn) is divisible by (a  b), if n is either odd or even.
What is the remainder when (1523 + 2323) is divided by 19? (CAT 2004, 2 marks)
It can be observed that (1523 + 2323) is divisible by 38, so it will be divisible by 19 also. Hence, remainder = 0.
Alternatively, this problem can be done either by cyclicity method or theorem method.

What is the remainder when (163 + 173 + 183 + 193) is divided by 70? (CAT 2005, 1 mark)
We know, this is a basic multiplication and division question. But using the above approach makes it a lot simple.
We know that (an + bn) is divisible by (a + b), if n is odd. Taking cue from this, we can say that (an + bn + cn) is divisible by (a + b + c), if n is odd and similarly (an + bn + cn + dn) is divisible by (a + b + c + d). Now 16 + 17 + 18 + 19 = 70, so remainder is zero. (As a, b, c, d are in AP)

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