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(i) If ‘a_{1}’ is divided by ‘n’, the remainder is ‘r_{1}’ and if ‘a_{2}’ is divided by ‘n’, the remainder is r_{2}. Then,
If a_{1}+a_{2} is divided by n, the remainder will be r_{1} + r_{2.}
If a_{1}  a_{2} is divided by n, the remainder will be r_{1}  r_{2.}
If a_{1} × a_{2} is divided by n, the remainder will be r_{1} × r_{2.}
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.
Example:
If 21 is divided by 5, the remainder is 1 and if 12 is divided by 5, the remainder is 2. Then, if (21 + 12 = 33) is divided by 5, the remainder will be (1 + 2 = 3).
If (21  12 = 9) is divided by 5, the remainder will be 1  2 =  1.
But if the divisor is 5,  1 is nothing but 4 (9 = 5 × 1 + 4)
So, if 9 is divided by 5, the remainder is 4 and 9 can be written as 9 = 5 × 2  1.
So here  1 is the remainder. So  1 is equivalent to 4 if the divisor is 5. Similarly  2 is equivalent to 3.
If (21 × 12 = 252) is divisible by 5, the remainder will be (1 × 2 = 2).
(ii) If two numbers ‘a_{1}’ and ‘a_{2}‘ are exactly divisible by n. Then their sum, difference and product is also exactly divisible by n.
i.e., If ‘a_{1}’ and ‘a_{2}’ are divisible by n, then
a_{1} + a_{2} is also divisible by n
a_{1}  a_{2} is also divisible by n
a_{1} × a_{2} is also divisible by n.
Example: 12 is divisible by 3 and 21 is also divisible by 3
Sol. So, 12 + 21 = 33, 12  21 =  9 and 12 × 21 = 252 all are divisible by 3.
Finding Remainders Of Powers With The Help Of Remainder Theorem
Example 1: What is the remainder if 7^{25} is divided by 6?
Solution: If 7 is divided by 6, the remainder is 1. So if 7^{25} is divided by 6, the remainder is 1 (because 7^{25} = 7 × 7 × 7… 25 times. So remainder = 1 × 1 × 1…. 25 times = 1^{25}).
Example 2: What is the remainder, if 3^{63} is divided by 14.
Solution: If 3^{3} is divided by 14, the remainder is  1. So 3^{63} can be written as (3^{3})^{21}.
So the remainder is ( 1)^{21} =  1.
If the divisor is 14, the remainder  1 means 13. (14  1 = 13) by pattern method.
The binomial expansion of any expression of the form
(a + b)^{n} = ^{n}C_{o} a^{n} + ^{n}C_{1} a^{n1} × b^{1} + ^{n}C_{2} × a^{n2} × b^{2} ..... + ^{n}C_{n1} × a^{1} × b^{n1} + ^{n}C_{n} × b^{n}
Where ^{n}C_{o}, ^{n}C_{1}, ^{n}C_{2}, .... are all called the binomial coefficients
In general, ^{n}C_{r} = n!/r!(n  r)!
There are some fundamental conclusions that are helpful if remembered:
(a) There are (n + 1) terms.
(b) The first term of the expansion has only a.
(c) The last term of the expansion has only b.
(d) All the other (n  1) terms contain both a and b.
(e) If (a + b)^{n} is divided by a, then the remainder will be b^{n} such that b^{n} < a.
Example 1: What is the remainder if 7^{25} is divided by 6?
Solution.
(7)^{25} can be written (6 + 1)^{25}. So, in the binomial expansion, all the first 25 terms will have 6 in it. The 26th term is (1)^{25}. Hence, the expansion can be written 6x + 1. 6x denotes the sum of all the first 25 terms.
Since each of them is divisible by 6, their sum is also divisible by 6, and therefore, can be written 6x, where x is any natural number. So, 6x + 1 when divided by 6 leaves the remainder 1.
(OR)
When 7 divided by 6, the remainder is 1. So when 7^{25} is divided by 6, the remainder will be 1^{25} = 1.
If n is a prime number, (n  1)! + 1 is divisible by n.
Lets take n = 5
Then (n  1)! + 1 = 4! + 1 = 24 + 1 = 25 which is divisible by 5.
Similarly If n = 7
(n  1)! + 1 = 6! + 1 = 720 + 1 = 721 which is divisible by 7.
Corollary
If (2p + 1) is a prime number (p!)^{2} + ( 1)^{p} is divisible by 2p + 1.
For Example
If p = 3, 2p + 1 = 7 is a prime number
(p!)^{2 }+ ( 1)^{p} = (3!)^{2} + ( 1)^{3} = 36  1 = 35 is divisible by (2p + 1) = 7.
PROPERTY
If “a” is natural number and P is prime number then (a^{p}  a) is divisible by P.
Example: If 2^{31} is divided by 31 what is the remainder?
So remainder = 2
FERMAT’S THEOREM
If p is a prime number and N is prime to p, then N^{p 1}  1 is a multiple of p.
Corollary
Since p is prime, p  1 is an even number except when p = 2.
Therefore ( ) = M(p).
Hence either 1 is a multiple of p, that is = Kp ± 1, where, K is some positive integer.
BASE RULE AND CONVERSION
This system utilizes only two digits namely 0 & 1 i.e. the base of a binary number system is two.
Example: 1101_{2} is a binary number, to find the decimal value of the binary number, powers of 2 are used as weights in a binary system and is as follows:
1 × 2^{3} = 8
1 × 2^{2} = 4
0 × 2^{1} = 0
1 × 2^{0} = 1
Thus, the decimal value of 1101_{2} is 1 × 2^{3} + 1 × 2^{2} + 0 × 2^{1} + 1 × 2^{0} = 13.
CONVERSION FROM DECIMAL TO OTHER BASES
We will study only four types of Base systems,
Let us understand the procedure with the help of an example
Example 1. Convert 357_{10} to the corresponding binary number.
Solution. To do this conversion, you need to divide repeatedly by 2, keeping track of the remainders as you go. Watch below:
As you can see, after dividing repeatedly by 2, we end up with these remainders:
These remainders tell us what the binary number is. Read the numbers outside the division block, starting from bottom and wrapping your way around the righthand side and moving upwards.
Example 2. (357)_{10} convert to (101100101)_{2}.
This method of conversion will work for converting to any nondecimal base. Just don't forget to include the first digit on the left corner, which is an indicator of the base. You can convert from baseten (decimal) to any other base.
CONVERSION FROM OTHER BASES TO DECIMAL
We write a number in decimal base as
345 = 300 + 40 + 5 = 3 × 10^{2} + 4 × 10^{1} + 5 × 10^{0}
Similarly, when a number is converted from any base to the decimal base then we write the number in that base in the expanded form and the result is the number in decimal form.
Example 1. Convert (1101)_{2} to decimal base
Solution. (1101)_{2} = 1 × 2^{3} + 1 x 2^{2} + 0 × 2^{1} + 1 × 2^{0} = 8 + 4 + 1 = 13
So (1101)_{2} = (13)_{10}
Example 2. Convert the octal no 3456 in to decimal number.
Solution. 3456 = 6 + 5 × 8 + 4 × 8^{2} + 3 × 8^{3}
= 6 + 40 + 256 + 1536
= (1838)_{10}
Example 3. Convert (1838)_{10} to octal.
Solution.
= (3456)_{8 }
Example 5. Find the fifth root of (15AA51)_{19.}
Solution. (15AA51)_{19} = 1.19^{5}+ 5.19^{4}+ 10.19^{3}+ 10.19^{2}+ 5.19^{1}+ 1.19^{0}
= (19+1)^{5} = 20^{5 } (Using binomial theorem)
Therefore, the fifth root is 20
IMPORTANT POINTS
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