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
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
n could be any of the integers. That's what the equations tell us at the end, that is, x = 57 ( -2000 + 22n ) ----- (1) y = 22 ( 5200 - 57n) ----- (2) where n belongs to the set of integers, Z = {...,-2, -1, 0, 1, 2,...} That's how we can get infinite pairs of (x,y) to solve this linear diophantine equation. Hope that helps.
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 .
Thats why you failed to find general solutions for this question. Nothing really complicated and Ive given out a simple formula you can use to come up with general solutions for linear diophantine equation. Not every math can be done with 8th grade level math bro. If you are not enough, right attitude is to LEARN and not complain anything as it is from your lack of understanding. Inventive thinking is something people like you should not be mentioning or just be quiet
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
Nice work! Thanks for sharing your work Devon👍👍👍
5x=400-143
=257/5
257+143/5=400
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Great, thank you Sir
Thanks a lot my friend👍👍👍
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
Nice👍👍👍
Ingenious method. thank you for your dazzling explanation sir
Thank you my man👍👍👍
Sir here how did you wrote it, explanation please 4:59
A= { a | 0 < a < 1 }
Therefore n(A) = ∞
Thats right sir haha👍👍👍
@@drpkmath12345 can you prove sir how it is ?
Nice technique!
Thank you for your comment mate👍👍👍
@@drpkmath12345 You are most welcome!
X=2-22k
Y=13+57k
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Very nice sir. Muchas gracias
De nada haha👍👍👍
Heyy sir i am From India 🇮🇳... (Kerala)
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@@drpkmath12345 hey how are u sir
@@Bluebirdgirl Hey how are you? Did you subscribe to my new channel?
i found x=22k+12 and y=(-627k-142)/11 using modular arithmetic
Nice work! Thanks for sharing my friend👍👍👍
How to find value of n?
n could be any of the integers.
That's what the equations tell us at the end, that is,
x = 57 ( -2000 + 22n ) ----- (1)
y = 22 ( 5200 - 57n) ----- (2)
where n belongs to the set of integers, Z = {...,-2, -1, 0, 1, 2,...}
That's how we can get infinite pairs of (x,y) to solve this linear diophantine equation.
Hope that helps.
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
Come to my new channel friend
Thank you sir
Come to my new channel my friend👍👍👍
Did you say x and y must be positive integers?
No
X=2,y=13
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What is "n"?
"n" is any integer.
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I am the first today
Thank you my friend👍👍👍
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 .
Thats why you failed to find general solutions for this question. Nothing really complicated and Ive given out a simple formula you can use to come up with general solutions for linear diophantine equation. Not every math can be done with 8th grade level math bro. If you are not enough, right attitude is to LEARN and not complain anything as it is from your lack of understanding. Inventive thinking is something people like you should not be mentioning or just be quiet