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
You may also like:
Who is behind QuickerMaths?
I am Vineet Patawari - PGDM (IIM Indore), ACA, B.Com(H). My passion for Mathematics, specially Vedic Maths encouraged me to start QuickerMaths
I believe that if trained properly using powerful tools like Vedic Maths, the immense intellect of human mind can be ignited instantly - find out more
Get Maths Tricks by Email
Recent Comments
its very nice n useful – 08Feb12
A and B can do a piece of work in 16days. B and C can... – 08Feb12
A, B and C can do a piece of work in 40, 60 and 120... – 07Feb12
as a night watchman he is not supposed to sleep during the night . as... – 06Feb12
ramlal is a night watchman and he is sleep on duty therefor boss fire him,... – 06Feb12
Most Popular
- 100% Finding Cube Root – Vedic Maths Way
- 67% Shortcut to Find Square of a Number
- 57% Shortcut Method for Multiplication
- 56% Shortcut to find the Cube of a number
- 52% Squaring any 2-digit number
- 49% Divisibility Rule of 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
- 45% Division in Vedic Mathematics
- 45% Herons Method of Finding Roots
- 43% You can make a difference
- 41% Quicker Maths by M Tyra
- 40% Quick calculations for extremely large numbers
- 39% Divisibility Rules for 7 , 11 and 13
- 37% How to convert from decimal to other number systems
- 36% Quick Multiplication up to 20 x 20
- 35% Logical Reasoning Books
April 27th, 2011 - 17:10
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
July 21st, 2011 - 18:27
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…
April 27th, 2011 - 17:08
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.
April 1st, 2011 - 22:03
how can i solve thisfind the remainder 3^50 divided by 11
April 2nd, 2011 - 10:30
3^50 = (3^5)^10= 243^10
When 243 is divided by 11 remainder is 1. Hence 1^10 = 1 >>Answer
Vineet Patawari
February 18th, 2011 - 14:27
what will b the remainder when N is divided by 11?
N=30 raised to the power 72 and 72 is raised to the power 87
August 2nd, 2011 - 22:18
answer is 4
May 17th, 2010 - 11:58
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
May 15th, 2010 - 08:39
Keep posting stuff like this i really like it
May 12th, 2010 - 19:57
Hi Vineet,
Why are you not replying to my query. Please I am hoping the answer from you….
May 12th, 2010 - 21:48
@ 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
May 7th, 2010 - 01:10
great post as usual!
May 6th, 2010 - 10:47
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
May 3rd, 2010 - 11:54
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.
April 30th, 2010 - 15:24
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
May 1st, 2010 - 19:01
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
April 30th, 2010 - 13:51
Mr. Vineet,
Thanks a lot for the help.
The following portion is not clear.
—————-
using remainder theorem
=(-1^4*16)/(641*625)
—————
Nandeesh
April 30th, 2010 - 10:41
How to find the remainder when 2^32 (2 raised to 32) is divided by 641?
April 30th, 2010 - 11:49
@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.
April 30th, 2010 - 10:27
wow i liked this way of solving questions
April 30th, 2010 - 11:50
@Chetna – Thanks for the appreciation.