Remainder Theorem for Number System | Remainder Theorem Difficult Examples

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Remainder Theorem of Numbers | Remainder Problems in Aptitude

Before going to concepts of remainder theorem of numbers, it is better to understand the the concepts of Divisor, Dividend, Quotient and Remainder

Remainder theorem of numbers | Remainder theorem aptitude questions | How to find remainder of a number with power | Remainder theorem tricks for ssc

Remainder Theorem Rule – 1 (Fundamental)

Remainder of the expression can be expressed as positive remainders and negative remainders. Technically both positive and negative remainders are correct.

But remainders by definition are always non-negative. So final answer to be expressed in positive value only. For simplification of solution take negative remainder and finally it is to be convert into positive remainders.

Note:

Add divisor to negative remainder then it will gives positive remainder
Subtract positive remainder from divisor then it will gives negative remainder

Example – 1 : Find the remainder of the expression of 49 / 9

Remainder theorem of numbers | Remainder theorem aptitude questions | How to find remainder of a number with power | Remainder theorem tricks for ssc

In this question

Positive remainder = +4

Negative remainder = -5

As per the remainder theorem the final answer is “4 “

Example – 2 : Find the remainder of the expression of 107 / 9

Remainder theorem of numbers | Remainder theorem aptitude questions | How to find remainder of a number with power | Remainder theorem tricks for ssc

In this question

Positive remainder = +8

Negative remainder = -1

In this expression finding the negative remainder is very easy when compare to positive remainder.

So take negative remainder “-1″ and add to it divisor ” 9″ then gives final answer = – 1 + 9 = 8

Remainder Theorem Rule – 2 (For long expressions)

The remainder of the expression $\frac{\left [ A \ \times B + \ C \ \times D \right ]}{N}$ will be the same as the remainder of the expression

$\frac{\left [ A _{R} \ \times B _{R} + \ C _{R} \ \times D _{R} \right ]}{N}$

Where

A_R is the remainder when “A” is divided by ” N”

B_R is the remainder when “B” is divided by ” N”

C_R is the remainder when “C” is divided by ” N”

D_R is the remainder when “D” is divided by ” N”

Example – 3 : Find the remainder when 47 x 52 is divided by 6

Solution: Remainder theorem of numbers | Remainder theorem aptitude questions | How to find remainder of a number with power | Remainder theorem tricks for ssc Here remainder for 47/6 is -1 or 5 and 52/6 is -2 or 4

Case – 1 : -1 x -2 = 2/ 6 $\rightarrow$ remainder = 2

Case – 2 : –1 x 4 = -4 / 6 $\rightarrow$ remainder = – 4 + 6 (add divisor) = 2

Case-3 : 5 x -2 = -10 / 6 $\rightarrow$ remainder = – 4 + 6 (add divisor) = 2

Case – 4 : 5 x 4 = 20 / 6 $\rightarrow$ remainder = 2

As per the reduces the calculation we can choose the option

Remainder Theorem Rule – 3 (Cancellation rule)

For simplification of the expression of the sum, you should try to cancel out parts of the numerator and denominator as much as you can, then final remainder to be multiplying the canceled number to get the correct remainder.

Example – 4 : Find the remainder of 54 divided by 4.

= 54/4 = 27 / 2 ( cancel out by 2 of the numerator and denominator )

Then remainder is of the final expression is 1

Final reminder for the given expression is = 1 x 2 = 2

Remainder Theorem Rule – 4 (remainder of a number with power )

There are two rules which are effect in order to deal with large powers

Case – 1 : If we can express the expression in the form $\frac{(qx \ + 1)^n}{q}$ , the remainder will become 1 directly.

In this case , there is no matter how large the value of the power “n” is, the remainder is 1.

Example – 5: Find the remainder when (321)⁵⁶⁸⁷ is divided by 8

Solution: 321 can be expressed as [(8×40) + 1] so

remainder of the above question is (1)⁵⁶⁸⁷ = 1

Case – 2 : If we can express the expression in the form $\frac{(qx \ - 1)^n}{q}$ , the remainder will become ( – 1 )ⁿ

In this case , if n is even number then remainder will be 1 and if n is odd number then remainder will be (q-1)

Example – 6 : Find the remainder when (146)⁵⁶ is divided by 7

Solution: 146 can be expressed as [(7×21) – 1] so

Remainder of the above question is (- 1)⁵⁶ = 1

Example – 7 : Find the remainder when (269)⁵⁷⁵⁸⁷ is divided by 6

Solution: 269 can be expressed as [( 6x 45) – 1] so

Remainder of the above question is (- 1)⁵⁷⁵⁸⁷ = – 1 = -1 + 6 = 5

Remainder theorem problems and solutions

Example – 8 : Find the remainder when 73 x 75 x 78 x 57 x 194 is divided by 25

Solution: $= \frac{ 73 \ \times 75 \times \ 57 \ \times\ 194657}{25}$

$= \frac{ -2 \ \times 0 \times \ 7 \ \times\ 7 }{25}$ = 0

Remainder theorem shortcut tricks : If denominator is perfectly divisible by any one number of given numerator expression values then remainder of the hole expression is Zero.

Example – 9 : Find the remainder when 84 + 98+ 197 + 240 + 140 is divided by 32

Solution: $= \frac{ 84 + 98 + 197 + 240 +140 }{ 32 }$