Find the remainder – Vedic Algebra

Vedic Algebra

I have got a mail from some QuickerMaths follower, to illustrate usage of Vedic Mathematics in branches of mathematics other than arithmetic. This post is for that purpose only. Here I am highlighting the usage of Vedic Mathematics in finding out the remainder when an algebraic expression is divided by another.

Finding out the remainder becomes extremely easy using Vedic Maths.

So lets begin with a simple example –


Find the remainder when

x3 + 4x2 + 6x – 7 is divided by (x + 5)

Solution:

Step I: Put divisor equal to 0 .i.e.

x + 5 = 0

x = -5

Step II: The remainder will be f(x).

f (-5) = (-5)3 + 4(-5)2 + 6(-5) – 7

= -125 + 100 – 30 – 7

= -62

Example 2: (Mx3 + 3x2 -3) and (2x3 – 5x +M) leaves the same remainder when divided by (x -4) Find the value of M.

Solution:

Let R1 and R2 be remainder for 1st and 2nd equation simultaneously

Rl = f (4) = M (4)3 + 3(4)2 – 3 = 64M+ 45

R2 = f (4) = 2(4)3 – 5(4) + M = M + 108

They leave the same remainder. So,

Since, Rl = R2. We have

64 M + 45= M +108

Or, 63 M = 63

M = 1

Example 4: (Mx3 + x2 – 2x – N) is exactly divisible by (x – 1) and (x + 1). Find the value of M and N.

Soln: When the expression is exactly divisible by any divisor, the remainder will be zero.

Now, the remainder, when the divisor is x-1, is

f (l) = M + 1 – 2 – N = 0          .

\M – N = 1                 ………….(1)

And the remainder, when the divisor is x + 1, is

f( -1) = M( -1)3 + (-1)2 – 2(-1) – N = 0

-M + 1 + 2 – N = 0

M + N = 3                     …..(2)

Solving (1) & (2), we have,

M = 2, N = 1

Thanks to Nehul from Nagpur for asking this question. If you have any similar question, go to Contact Page and post your queries/suggestions.

Take Care. God Bless!!

Vineet Patawari

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Vineet Patawari

Hi, I'm Vineet Patawari. I fell in love with numbers after being scared of them for quite some time. Now, I'm here to make you feel comfortable with numbers and help you get rid of Math Phobia!

28 thoughts to “Find the remainder – Vedic Algebra”

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  4. Hi, could you please tell me how to solve this problem? A number when divided by 32 leaves a remainder of 29. Find the remainder when the number is divided by 64

    1. Hi Pallavi!!!
      I am too late to see your problem but any ways;
      the answer is either 29 or 0.
      If the multiplicand of 32 is odd, the remainder will be 32 otherwise it will be 0…
      Hope to hear from you soon….
      Have a nice time…

  5. Hi, could you please tell me how to solve this problem? A number when divided by 32 leaves a remainder of 29. Find the remainder when the number is divided by 64.

  6. Hi Vineet,

    Thanks very much for your answer to my question. You are really proving your dedication over here. Thanks once again. I am sure that we can always hope more from you.

    But to be frank,I am still not getting the below step:

    How you are saying that ,

    16/(641*625) =(16-641)/(641*625)

    Based on which rule or theorem you are subtracting 641 from 16?

    Thanks in advance Vineet.
    — Rakesh

    1. @ Rakesh – Sorry for missing your question.

      I am trying my best to explain it over here –
      =16/(641*625)
      =(16-641)/(641*625)
      =-625/(641*625)
      =-1/641 (removing 625 from both numerator and denominator)
      Therefore remainder = 640

      Try to understand this by a simple example –
      21/144
      =(7*3)/(48*3)
      =21/144
      Remainder = 18

      now,
      7/48 — > remainder is -1 or (7-1) = 6

      So if you have a common factor in both numerator and denominator, you can remove that for make calculations simple.
      So if we remove 3, the problem looks like
      7/48
      in this case the remainder is -1
      the final answer has to be multiplied again by the common factor 3
      hence the remainder is -1*3 = -3
      or, 21-3 = 18

  7. There is still confusion,
    how did you applied Remainder theorem to the below step :

    =16/(641*625)

    =-625/(641*625)

    this form looks like (a/b*c) which does not fit to any of the above mentioned forms?

    Please mention exact solution

  8. Hi Vineet, Thanks for your response, I am really happy the way you are solving every question posted. Hats off to you. The video you have provided is really fantastic. And also your solution is now cleared all confusions that were in my mind. Thanks once again.

  9. Thanks Avinash, for quick response with the answer.
    But still I didnt the following step:

    =16/(641*625)

    =-625/(641*625)

    Also the step asked by Nandeesh is also not understood by me…

    —Nikhil Pawar

    1. I believe the main confusion is due to the basics of REMAINDER THEOREM –

      Remainder (a x b)/c = Remainder (a/c) x Remainder (b/c)
      Remainder (a + b)/c = Remainder (a/c) + Remainder (b/c)
      Remainder (a x b)/c = Remainder (a/c) – Remainder (b/c)
      Remainder (a / b)/c = Remainder (a/c) / Remainder (b/c)

      I am giving you a video link to understand the concept –
      http://www.mbatown.com/community/video/Tutorials-on-Quantitive-Aptitude-Maths-Problem-Solving-Number-System-Remainder-Theorem-1

  10. Mr. Vineet,
    Thanks a lot for the help.
    The following portion is not clear.

    —————-
    using remainder theorem

    =(-1^4*16)/(641*625)

    —————

    Nandeesh

    1. @Nikhil – Thanks for asking that question.

      2^32/641

      =(128^4 * 16)/641

      multiplying both numerator & denominator by 5^4 (since 5*128=640), we get

      (128^4 * 16*5^4)/(641*(5^4))

      =(640^4*16)/(641*(5^4))

      using remainder theorem

      =(-1^4*16)/(641*625)

      =16/(641*625)

      =-625/(641*625)

      =-1/641

      Therefore remainder is -1 or 641-1 = 640

      I have given the above solution without mod (I avoided that to remove the confusion & to keep it simple)

      If you have difficulty in understanding the above solution do let me know, I have an alternative solution also.

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