A Nice Diophantine Equation in Number Theory | You Should Learn This Theorem | Math Olympiad

In this video, I am introducing a nice diophantine equation in number theory and also a wonderful theorem you can use to solve a linear diophantine equation with two unknown variables. With this theorem, most of the linear diophantine equations with two unknown variables will be nicely solved. Diophantine equation is an interesting topic in number theory. Come check this video out and watch it until the end. This would also be a good practice for math olympiad. More with diophantine equation problems will come! Stay tuned!

💬 Did you enjoy this video? Tell me all about it in the comment section below!

Subscribe to Dr. PK Math here ⤵️

*************
✅ Follow Modern Dr PK Math on social media:

*************
#DrPKMath #diophantineequation #matholympiad

[devondevon]

22y + 57x =400
~ this will use for congruent to
0y + 13x `~ 4 ( mod 22)
13 x ~26 (mod 22)
x ~ 2 (mod 22)
x = 2 + 22k equation A
Since 22y + 57x =400, then
22y + 57 ( 2+22k) =400
22y + 114 + (57)(22k) =400
22y = 286 - (57)(22k)
y = 13- 57k equation B
when k= 1 , x = 2+22 = 24 and
when k=1, y = 13-57 = -44
So one solution is (24 - 44)
let's plug in these values in the original equation, 22y + 57k=400
22(-44) + 57(24) = 400
-968 + 1368 =400
400 = 400
Trying other values when k=2
x= 2+ 44 = 46 and
y= 13-114 = - 101
let's plug in 46 and - 101 into the original equation
(22)(-101) + 57(46 = 400
-2, 222 + 2622 = 400
400 = 400
So, the solution for the linear diophantine equation 57x + 22y =400
is x= 2+22k, and y =13-57k

[nasrullahhusnan]

mod(57x, 2)=0 --> x is even
x=2k where k is positive integer
Thus the equation may written as
57×2k+22y=400 --> 57k+11y=200
As the last digit of RHS is 0 then sum of last digit of 57k and 11y must be 0. Note that k<4 and y<20 Thus
(k, y)={(1, 3), (1, 13), (2, 6), (2, 16), (3, 9), (3, 19)}
For (k, y)=(1, 13), 57k+11y=57+143
=200

[sarahsiddiqui]

Sir here how did you wrote it, explanation please 4:59

[MrGLA-zsxt]

Ingenious method. thank you for your dazzling explanation sir

[iiwacky]

i found x=22k+12 and y=(-627k-142)/11 using modular arithmetic

[Crazy_mathematics]

A= { a | 0 < a < 1 }

Therefore n(A) = ∞

[richardl]

Did you say x and y must be positive integers?

[Bluebirdgirl]

Heyy sir i am From India 🇮🇳... (Kerala)
Nice class🎉🙌

[satyapalsingh]

x=2+22k, k=0,1,2,3,……..,y=13-57k, k=0,1,2,3,…………..,(x,y)=(2,13),(24,-44),(46,-101),……………Answer

[arunachalamhariharan]

This method is just too much use of higher mathematics .
I use only school level ( 8th class level ) mathematics to solve such simple problems .
The core of my thinking is
400 = ( 7 × 57 ) + 1
&
( 57 × n ) + 1 = 22 × p
Finding n and p is a school level thinking .

My request
PLEASE DO NOT COMPLICATE SIMPLE SCHOOL LEVEL THINKING .
THIS WILL KILL INVENTIVE THINKING .
Thanks .