Table of contents  
Introduction  
Concept of Remainder Theorem  
Prime Number Divisor Rule  
Different Types of Questions  
Some Special Cases 
Supposing a number(or dividend) "D" is divided by another number “d"(divisor); if the quotient obtained is “Q” and the remainder obtained is “R”, then the number can be expressed as D = Q(d) + R.
Dividend = Quotient(Divisor) + Remainder 
Example: Suppose 8 is divided by 3.
In this case, N = 8, x = 3; 3 × 2 = 6, which is 2 less than 8.
Hence, Q = 2 and R = (8  6) = 2
Hence, 8 = 2 × 3 + 2
The remainder theorem is based on the product of individual remainders. If R is the remainder of an expression (p*q*r)/X, and p_{R}, q_{R} and r_{R} are the remainders when p, q and r are respectively divided by X, then it can be said that ((p_{R} × q_{R} × r_{R} ))/X, will give the same remainder as given by (p*q*r)/X
Let us understand this with the help of an example:
Example 1: Find the remainder when (361*363) is divided by 12.
Ans:
 Take the product of individual remainders, i.e. 361/12  R =1 and 363/12  R= 3.
 Find the remainder when you divide that product by the number (361*363)/12  R= (1*3)/12  R. => Answer = 3
 This is Basic Remainder theorem put across in Numbers.
Example 2: Find the remainder when 10^{6} is divided by 7, i.e. (10^{6}/7)R.
Ans: 10^{6} can be written in the form of 10^{3} x 10^{3}
 10^{6 }= 10^{3 }x 10^{3}
 Thus (10^{6}/7)R = (10^{3}/7 x 10^{3}/7)R = ((6 * 6)/7)R = (36/7)R = 1.
 So the remainder is 1.
By definition, remainder cannot be negative. But in certain cases, you can assume that for your convenience. But a negative remainder in real sense means that you need to add the divisor in the negative remainder to find the real remainder.
Let us see why this happens:
If the numbers N1, N2, N3 give remainders of R1, R2, R3 with quotients Q1, Q2, Q3 when divided by a common divisor D
► N1 = DQ1 + R1 N2 = DQ2 + R2 N3 =DQ3 + R3
► Multiplying = N1 * N2 * N3
= (DQ1 + R1) * (DQ2 + R2) * (DQ3 + R3) = D(some number) + (R1 * R2 * R3)
= First part is divisible by D
Hence, you need to check for the individual remainders only.
The "prime number divisor rule" is a concept related to prime numbers and their divisors. It states that : If P is a prime number, then : The remainder of the expression A^{(P1)} / P is 1.
Example 1: What is the remainder when the product 1998 × 1999 × 2000 is divided by 7?
Ans: The remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively.
 Hence the final remainder is the remainder when the product 3 × 4 × 5 = 60 is divided by 7. Therefore, the remainder = 4
Example 2: What is the remainder when 2^{2004} is divided by 7?
Ans: 2^{2004 }is again a product (2 × 2 × 2... (2004 times)).
 Since 2 is a number less than 7, we try to convert the product into the product of numbers higher than 7.
 Notice that 8 = 2 × 2 × 2.
 Therefore we convert the product in the following manner 2^{2004} = 8^{668 }= 8 × 8 × 8... (668 times).
 The remainder when 8 is divided by 7 is 1.
 Hence the remainder when 8^{668} is divided by 7 is the remainder obtained when the product 1 * 1 * 1... is divided by 7
Answer = 1
Example 3: What is the remainder when 2^{2006} is divided by 7?
Ans: This problem is like the previous one, except that 2006 is not an exact multiple of 3 so we cannot convert it completely into the form 8^{x}.
 We will write it in the following manner: 2^{2006} = 8^{668} × 4.
 Now, 8^{668} gives the remainder 1 when divided by 7 as we have seen in the previous problem.
 And 4 gives a remainder of 4 only when divided by 7.
 Hence the remainder when 2^{2006} is divided by 7 is the remainder when the product 1 × 4 is divided by 7. Therefore, the remainder = 4
Example 4: What is the remainder when 25^{25} is divided by 9?
Ans: Again 25^{25} = (18 + 7)^{25} = (18 + 7)(18 + 7)...25 times = 18K + 7^{25}
 Hence remainder, when 25^{25} is divided by 9, is the remainder when 7^{25} is divided by 9.
 Now 7^{25} = 7^{3} × 7^{3} × 7^{3}.. (8 times) × 7 = 343 × 343 × 343... (8 times) × 7.
 The remainder when 343 is divided by 9 is 1, and the remainder when 7 is divided by 9 is 7.
 Hence the remainder when 7^{25} is divided by 9 is the remainder we obtain when the product 1 × 1 × 1... (8 times) × 7 is divided by 9.
 The remainder is 7 in this case. Hence the remainder when 25^{25 }is divided by 9 is 7.
Hence to find the remainder when both the dividend and the divisor have a factor in common:
Example : What is the remainder when 3^{444} + 4^{333} is divided by 5?
Ans: The dividend is in the form a^{x} + b^{y} . We need to change it into the form a^{n} + b^{n}.
 3^{444} + 4^{333} = (3^{4}) ^{111} + (4^{3}) ^{111}.
 Now (3^{4}) ^{111} + (4^{3}) ^{111} will be divisible by 3^{4} + 4^{3 }= 81 + 64 = 145.
 Since the number is divisible by 145, it will certainly be divisible by 5.
Hence, the remainder is 0.
Example: Find the remainder when x^{3}  ax^{2} + 6x  a is divided by x  a.
Ans: Let p(x) be any polynomial of degree greater than or equal to one and let 'a' be any real number. If a polynomial p(x) is divided by x  a then the remainder is p(a).
Let p(x) = x^{3}  ax^{2} + 6x  a
The root of x  a = 0 is a.
p(a) = (a)^{3}  a(a)^{2} + 6(a)  a
= a^{3}  a^{3} + 5a
= 5a
Hence by remainder theorem, 5a is the remainder when x^{3}  ax^{2} + 6x  a is divided by x  a.
Example: If 2x ^{3 }3x ^{2 }+ 4x + c is divisible by x – 1, find the value of c.
Ans: Since the expression is divisible by x – 1, the remainder f(1) should be equal to zero.(by Remainder Theorem)
Or 2 – 3 + 4 + c = 0, or c = 3.
If p is a prime and a is a positive integer
a) For a p and any integer a,
a^{p }≡ a(mod p)
b) Furthermore, if p ∤ a,
then a^{p}^{−}^{1}≡1(mod p)
For example, if a = 2 and p = 7, then 2^{7} = 128, and 128 − 2 = 126 = 7 × 18 is an integer multiple of 7.
Example: What is the remainder when n^{7} – n is divided by 42?
Ans: Since 7 is prime, n^{7} – n is divisible by 7.
 n ^{7 }– n = n(n ^{6 }– 1) = n (n + 1)(n – 1)(n ^{4 }+ n ^{2 }+ 1)
 Now (n – 1)(n)(n + 1) is divisible by 3! = 6
 Hence n ^{7 }– n is divisible by 6 x 7 = 42.
 Hence the remainder is 0.
If p is a prime number, then p divides (p−1)!+1.
Example: Find the remainder when 16! Is divided by 17.
Ans: 16! = (16! + 1) 1 = (16! + 1) + 16 – 17
 Every term except 16 is divisible by 17 in the above expression.
 Hence, the remainder = the remainder obtained when 16 is divided by 17 = 16
Example: How many numbers between 1 and 400, both included, are not divisible either by 3 or 5?
Ans: We first find the numbers that are divisible by 3 or 5. Dividing 400 by 3 and 5, we get the quotients as 133 and 80 respectively.
 Among these numbers divisible by 3 and 5, there are also numbers which are divisible both by 3 and 5, i.e. divisible by 3 x 5 = 15.
 We have counted these numbers twice. Dividing 400 by 15, we get the quotient as 26.
 Hence, the number divisible by 3 or 5 = 133 + 80 – 26 = 187
 Hence, the numbers not divisible by 3 or 5 are = 400 – 187 = 213.
184 videos131 docs110 tests

1. What is the concept of the Remainder Theorem? 
2. How is the Prime Number Divisor Rule related to the Remainder Theorem? 
3. What are the different types of questions that can be solved using the Remainder Theorem? 
4. Are there any special cases to consider when applying the Remainder Theorem? 
5. How can the Remainder Theorem be applied to solve problems in competitive exams? 

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