Remainder Theorem | Basic fundamental approach.

Sohel Sahoo: Hello Guys,
 In this article ,We would learn the basic fundamental approach & applications of Remainder Theorem to solve out numerous problems asked in various competitive exams including CAT,CDS, SSC CGL, SSC CHSL, SSC CPO, BANK PO, BANK CLERK, NTSL,RAILWAY, LDC,CSAT etc. like any one day exam.
         

Note:-   This Remainder must be zero or positive(+ sign only) & less than divisor as given in final answer of exams.               

Sign:-  The Remainder obtained on dividing the Dividend by the Divisor is often represented by  R[Dividend/Divisor] .

Technical Approach of Remainder :-

As you saw above that there are two remainders viz. Positive & Negative obtained in the process of division.

• You  normally take  a number just smaller than or equal to the Dividend,which is divisible by the divisor and while subtracting it from the Dividend;you get Positive Remainder or simply zero (in perfectly divided case).

• In a similar manner you can also take a number just greater than the Dividend,which is divisible by the divisor and on subtracting it from the Dividend;the result will be Negative Remainder.

Observation:-
      (1) Numeric  [ Excluding the signs + & -] Sum of both Remainders is equal to the Divisor.
                       e.g. R[19/6] =  1 & -5 . Here 1+5 = 6 (Divisor)
           Therefore, if you have one remainder (+/-);you can get other by subtracting it from the divisor & just putting an oposite sign (-/+) before it.
             In above example the positive remainder is 1 ; so, the negative remainder will be -(6-1)=-5 .
But, if you have negative remainder then just add the divisor to get the positive remainder.
    In above -ve remainder is -5 .so, +ve remainder = +(9-5)= -5+9=4

(2) Negative Remainder is also technically correct but in exams the given answer options are always positive.So, we can use Negative Remainer  for calculation purpose. Take that Remainder which is smaller in numeric value for easy calculation.

 Concepts:-
(1.)     R[30/9] = 3 or -6
In case we simplify 30/9 on cancelling by 3 we have 10/3 .
       Now, R[10/3]= 1 or -2 ; this is also cancelled by 3.
So, inorder to attain the original value  of Remainders we have to multiply by 3 in Remainders  obtained after cancellation.

NOTE:- When you cancel the Numerator & Denominator by a factor;you have to multiply by that factor to the Remainder so obtained on Cancellation.
    
(2.)  '0' is the smallest number which is divisible by any number.
         e.g. R[11/17] = 11 or -6
The reason behind this is 17 x 0 = 0 ; 0 is the smaller no. to dividend 11 here.
APPLICATION:-

While we are dealing with finding Remainders of an expression involving basically Addition or Multipliction of certain numbers and their equal range forms such as Subtraction,Division,Higher powers,Factorial notation ; we can find Remainders of individual number with the divisor and carry out the expression as it is. In the end, the remainder so obtained should be less than divisor unless divide again to find out so. [ If you want to know the reason behind this then comment down for which I shall explain it to you & so do I. ]

Eg: Find remainder of  (8 x 5 x 2) / 3 .

Sol:- Dividing individually each no.s by 3 we have remainders {-1/2},{2/-1},{2/-1}.
       By taking any remainder we have same result as explained below.
  1. (-1) x (-1) x (-1)= -1 = -1+3=2 (Ans)
  2. 2 x 2 x2 =8 =>R[8/3]=2
 3. 2 x (-1) x (-1) = 2
That's the reason why we obtain the same result irrespective of the remainder chosen;hence proving the authenticity of both +ve&-ve remainders in every such cases.
Q. Find remainder of (66 x 79) / 9.
Sol:-  R[66/9] = 3 or -6
          R[79/9] = -2 or 7
   Hence, R[66 x 79/9] = 3(-2)= -6 = -6+9= 3.(Ans)
Q. What will be the remainder when 47 x 53 x 71 is divided by 25 ?
Sol:- R[47/25] = -3 or 22
         R[53/25]= 3 or -22
        R[71/25] = -4 or 21
Now,R[(47x53x71)/25] = (-3)(3)(-4) = 36 => R[36/25] = 11(Ans)
# If you solve this by basic division i.e. by taking positive remainder then this would consume more time & is quite complicated. Here,negative remainder helps a lot.
Q. Find remainder when [42x35x20] is divisible by 16.
Sol:- Directly, Placing individual remainders we have 10x3x4 = 120
                                                                                   => R[120/16]= 8
 By  Simplification,  (42 x 35 x 20)/ 16
                                 => (21 x 35 x 5)/2   [Simplified by 2&4 i.e. by 8]
        Now, replacing individual remainders we get,
                  1 x 1 x1 =1
But,we have simplified the expression in total by 8.So,the actual remainder will be 1 x 8= 8 (Ans)
Q. Find the remainder when 183+5412 - 2054 -151 -283+608 is divided by 5 .
Sol:-   Placing individual remainders we get,  [Apply divisibility rule of 5]
                  (-2)+2-(-1)-1-(-2)+3 = 5 =>R[5/5] = 0
FACTORIALS:-
 n factorial is often written as the product of first n natural numbers in decreasing manner.It is represented by 'n!' .

e.g. 3!=3x2x1
       4!=4x3x2x1
      6!=6x(5x4x3x2x1)= 6 x 5!
      10! = 10 x 9 x 8! and so on.
Q. Find the remainder when 1!+2!+3!+4!+.......+100! is divided by10.
Sol:-  Putting individual remainders in the expression we have,
         1!=1 & R[1/10] = 1 or -9
         2!= 2 & R[2/10] = 2 or -8
         3!= 6 & R[6/10]= 6 or -4
        4!=24 & R[24/10]= 4 or -6
         5!=120 & R[120/10]= 0 {you can also find out as 2&5 make 10 which is in 5!}
     As, every factorial greater than 5 contain 5!;hence all are divisible by 10 & leavws remainder 0.
    So, the expression becomes 1+2+(-4)+4+0+0+.....+0 = 3 (Ans)

             Hope,these concepts as well as techniques help you a lot to create your basics about finding remainder smartly which will in future help you to understand Remainder Theorems to solve more difficult problems.

 If you have any query then feel free to contact us & kindly give your valuable opinions in comment box for improvement of the present methods.And subscribe for getting new updates from this as it ain't over & continued.