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**FINDING REMAINDERS OF A PRODUCT (DERIVATIVE OF REMAINDER THEOREM)****(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 **

**Concept of Negative Remainder: **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.****Ex**. 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 ‘

i.e., If ‘

**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:**

**1. Example: ****What is the remainder if 7 ^{25} is divided by 6?**

**2. Example: ****What is the remainder, if 3 ^{63} is divided by 14.**

So the remainder is (- 1)

If the divisor is 14, the remainder - 1 means 13. (14 - 1 = 13) by pattern method.

**3. Example: Find remainder when 4 ^{33} is divided by 7.**

If 4

If 4

If 4

The remainders of the powers of 4 repeats after every 3rd power.

So, as in the case of finding the last digit, since the remainders are repeating after every 3^{rd} power, the remainder of 4^{33} is equal to the remainder of 4^{3} ( since 33 is exact multiple of 3) = 1.

(OR)

If 4^{3}^{3} is divided by 7, the remainder is 1. So 4^{33} = (4^{3})^{11} is divided by 7, the remainder is 1^{11} = 1.**APPLICATION OF BINOMIAL THEOREM IN FINDING REMAINDERS**

The binomial expansion of any expression of the form

(a + b)^{n} = ^{n}C_{o} a^{n} + ^{n}C_{1} a^{n-1} × b^{1} + ^{n}C_{2} × a^{n-2} × b^{2} ..... + ^{n}C_{n-1} × a^{1} × b^{n-1} + ^{n}C_{n} × b^{n}

Where ^{n}C_{o}, nC1, ^{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, i.e.**(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?**

(7)

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.

**2. Example: ****Remainder when 25 ^{10} is divided by 576?**

Please note that 576 = 24

There are couple of methods of solving this.

Using Binomial Theorem

25

In the expansion, there will be 11 terms where the powers of 24 will vary from 0 to 10.

If the power of 24 is greater than or equal to 2 in a term, that term will be divisible by 576

The terms that will not be divisible by 576 are the terms that have powers of 24 as 0 or 1.

Those terms are

10C

= 10.24.1 + 1.1.1

= 241

So, Remainder is [25

**Wilson’s Theorem****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.**CorollaryIf (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!)

**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,

- Binary system (0, 1)
- Octal system (0, 1, 2, 3, 4, 5, 6, 7).
- Decimal system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
- Hexa-decimal system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C D, E, F) where A = 10, B = 11 … ,F= 15.

__Let us understand the procedure with the help of an example__**Example 1. Convert 357 _{10} to the corresponding binary number.**

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 right-hand side and moving upwards.**(357) _{10} convert to (101100101)_{2}.**

This method of conversion will work for converting to any non-decimal 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 base-ten (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**

So (1101)

**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.**

= (3456)_{8}

**Example 4. ****What is the product of highest 3 digit number & highest 2 digit number of base 3 system?(1) (21000) _{3 }**

222 = 2 + 2 × 3 + 2 × 3

22 = 2 + 2 × 3 = 8

∴ Product = 26 × 8 = 208

Convert back to base

(21201)

**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**

**Example 6. ****In a number system the product of 44 and 11 is 1034. The number 3111 of this system, when converted to the decimal number system, becomes?****(1) 406****(2) 1086 ****(3) 213 ****(4) 691****(5) None of the above****Solution.**

Let the base be n

(4n+4)(n+1)= n^{3}+3n+4

⇒ n^{3}-4n^{2}-5n=0

⇒ n(n-5)(n+1)=0

⇒ n=5

⇒(3111)_{5 }= (406)_{10}**IMPORTANT POINTS**

- The sum of consecutive five whole numbers is always divisible by 5.
- The square of any odd number when divided by 8 will leave 1 as the remainder
- The product of any three consecutive natural numbers is divisible by 8.
- The unit digit of the product of any nine consecutive numbers is always zero.
- For any natural number n, 10
^{n}^{-7 }is divisible by 3. - Any three-digit number having all the digits same will always be divisible by 37.

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