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 pR, qR and rR are the remainders when p, q and r are respectively divided by X, then it can be said that ((pR × qR × rR ))/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 106 is divided by 7, i.e. (106/7)R.
Ans: 106 can be written in the form of 103 x 103
- 106 = 103 x 103
- Thus (106/7)R = (103/7 x 103/7)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(P-1) / 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 22004 is divided by 7?
Ans: 22004 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- 22004 = 8668 = 8 × 8 × 8... (668 times).
- The remainder when 8 is divided by 7 is 1.
- Hence the remainder when 8668 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 22006 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 8x.
- We will write it in the following manner: 22006 = 8668 × 4.
- Now, 8668 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 22006 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 2525 is divided by 9?
Ans: Again 2525 = (18 + 7)25 = (18 + 7)(18 + 7)...25 times = 18K + 725
- Hence remainder, when 2525 is divided by 9, is the remainder when 725 is divided by 9.
- Now 725 = 73 × 73 × 73.. (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 725 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 2525 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 3444 + 4333 is divided by 5?
Ans: The dividend is in the form ax + by . We need to change it into the form an + bn.
- 3444 + 4333 = (34) 111 + (43) 111.
- Now (34) 111 + (43) 111 will be divisible by 34 + 43 = 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 x3 - ax2 + 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) = x3 - ax2 + 6x - a
The root of x - a = 0 is a.
p(a) = (a)3 - a(a)2 + 6(a) - a
= a3 - a3 + 5a
= 5a
Hence by remainder theorem, 5a is the remainder when x3 - ax2 + 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,
ap ≡ a(mod p)
b) Furthermore, if p ∤ a,
then ap−1≡1(mod p)
For example, if a = 2 and p = 7, then 27 = 128, and 128 − 2 = 126 = 7 × 18 is an integer multiple of 7.
Example: What is the remainder when n7 – n is divided by 42?
Ans: Since 7 is prime, n7 – 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.
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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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