What we did in math for IT class we used extended Euclidean algorithm x = 1(mod 3) x = 2(mod 5) 1=3(2)+5(-1) x = (3*2*2+5*(-1)*1)(mod 15) x = 7 (mod 15) and in second step we have two congruences instead of three x = 7 (mod 15) x = 3 (mod 7) 1 = 15*(1) + 7*(-2) x = (15*1*3 + 7*(-2)*7 ) (mod 105) x = (45 - 98) (mod 105) x = (45 - 98 + 105)(mod 105) x = (45+7)(mod 105) x = 52 (mod 105)
Thank you It is envolve momery I would set x =3a+1 3a+1 =2 ( mod 5) 3a = 1 ( mod 5) The inverse of 3 is 2 a=2 ( mod 5 ) a = 5b +2 Plug a in x x= 3(5b +2)+1 = 15b +7 15b+6 = 3 ( mod 7) 15b = 3 ( mod 7) b = 3 ( mod 7) b = 7c +3 Plug b in x x = 15b+2 = 15(7c+3) +7 = 105c+ 45 + 7 = 105c + 52 Or x = 52 (mod 105)
This is a new concept to me, very good lesson! ❤ May i ask why it doesnt work if the m’s have a common factor? And if they do, is there a way to perhaps change the m so it doesnt share any factors?
that doesn't happen if all the moduli are coprime. so if it happens, try to adjust the non-coprime moduli, if that's not possible, then there's no valid solution to the system
12 = 2*2*3 -> factorizing modulus in primes x = 8 (mod 12) becomes: x = 8 (mod 2), and x = 8 (mod 2), and x = 8 (mod 3) note that the first two equations are the same so only need it once x = 8 (mod 2) = 0 (mod 2), and x = 8 (mod 3) = 2 (mod 3)
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thanks for making this video. your explanation is far better than my CS professor
You're an awesome explainer! Thanks a lot.💯
Oh my goodness, you're a savior...love you so much
What we did in math for IT class
we used extended Euclidean algorithm
x = 1(mod 3)
x = 2(mod 5)
1=3(2)+5(-1)
x = (3*2*2+5*(-1)*1)(mod 15)
x = 7 (mod 15)
and in second step we have two congruences instead of three
x = 7 (mod 15)
x = 3 (mod 7)
1 = 15*(1) + 7*(-2)
x = (15*1*3 + 7*(-2)*7 ) (mod 105)
x = (45 - 98) (mod 105)
x = (45 - 98 + 105)(mod 105)
x = (45+7)(mod 105)
x = 52 (mod 105)
nice but do u know what was the matrix method he was talking abt?
Very good teacher 🎉
What a great explanation!!!
Thank you
It is envolve momery
I would set x =3a+1
3a+1 =2 ( mod 5)
3a = 1 ( mod 5)
The inverse of 3 is 2
a=2 ( mod 5 )
a = 5b +2
Plug a in x
x= 3(5b +2)+1 = 15b +7
15b+6 = 3 ( mod 7)
15b = 3 ( mod 7)
b = 3 ( mod 7)
b = 7c +3
Plug b in x
x = 15b+2 = 15(7c+3) +7
= 105c+ 45 + 7
= 105c + 52
Or x = 52 (mod 105)
Thank you so much your videos were very helpful
Thanks very much sir I'm looking out for more
You're so helpful omg.
This is gold ! Thanks a lot !
This is a new concept to me, very good lesson! ❤
May i ask why it doesnt work if the m’s have a common factor? And if they do, is there a way to perhaps change the m so it doesnt share any factors?
The proof is built on the fact that the gcd of the m's is 1. In short. Most theorems in Number Theory are based on this. Someday, I'll do the proof.
No doubt remains that we need Prime Newtons! 😊🎉
I had it on math for IT class
And what if when finding b1 the combo doesn’t have an inverse. What do you do then?
that doesn't happen if all the moduli are coprime. so if it happens, try to adjust the non-coprime moduli, if that's not possible, then there's no valid solution to the system
you said we need to break the modulus down to where it has a prime value. how do we do that? (Ex. my question has 8mod12)
12 = 2*2*3 -> factorizing modulus in primes
x = 8 (mod 12) becomes:
x = 8 (mod 2), and
x = 8 (mod 2), and
x = 8 (mod 3)
note that the first two equations are the same so only need it once
x = 8 (mod 2) = 0 (mod 2), and
x = 8 (mod 3) = 2 (mod 3)
As far as Chinese inventions go the CRT ranks right along side Kung Pao Chicken AFAIAC. 😋
Had to Google Kung Pao Chicken 🤣
❤❤❤❤@@PrimeNewtonsI am an Olimpiad student in Ghana and I like your lessons a lot
👍👍👍👍👍👍👍👍👍👍👍👍very good
Great!
Thanks man
what does b and c represent
Did you watch the video?
@@PrimeNewtons i did, but I'm a bit slow lol
at 4 mins into the video 4:36
I thought if critical race theory when I saw CRT
man ❤
W video
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