15 = 5 mod 10 if we common out 5 then it will be 3 = 1 mod 2 and not 3 = 1 mod 10.... so getting answer correct at the end of video is just a coincidence .... in short we are right 4
You explained clearly sir... Keep putting new vedios.. Keep going up.. Continue rocking.. Put more vedios.. Continue I will always support you sir. .. Very soon you will reach more & more subscribers... 🎉 All the best...
In equation 2, we need to replace 2x = 6 mod 20 by x = 3 mod 10 for getting correct answer. (please note that 6 cannot be directly divided by 2. there will be change in mod part as well)
@5:38 , the equation written is wrong , it should be x= 3 mod10 , we also have to divide inside the mod, eg. 10=2mod8 implies 5=1mod4 , not 5=1mod8 , so the solution is wrong
1:45 yes we can, but what X is? what we are calculating. where those other come? like this is private key and this cipher and x answer is private key. like real example not just this numbers. what they are?. this was exact same video but use diff numbers that other. what have to do network security?
The given problem states that \( X = 122 \) for mod 3, 5, and 11. We can't get \( a1, a2, a3 \) as 5, 2, and 1, respectively. Therefore, the question is incorrect. The correct values for \( a1, a2, a3 \) should be 2, 2, and 1. hope you understand , if i am wrong than let me know the correction..
I went through the problem but fixed the error at 5:25 and got x = 53 (mod 90). Which works, but it would nice for the modulus to be 180. Here's what I did: x = 53 (mod 90) x = 90k + 53 for some integer k. k is either even or odd, in other words leaves a remainder of 0 or 1 when divided by 2. If k is even: x = 90k + 53 = 90(2m) + 53 = 180m + 53, where m is another integer. So, x = 53 (mod 180) is a solution. If k is odd: x = 90k + 53 = 90(2n + 1) + 53 = 180n + 90 + 53 = 180n + 143, where n is an integer. So, x = 143 (mod 180) is the second solution.
HW: X = 122, but when 122 is divided by 3, we get 2 as the remainder, not 5 Which doesn't fit for x = 5 (mod 3). Or is X = 5 (mod 3) equivalent to X = 2 (mod 3) since 5 / 3 = r2?
its correct. 5(mod 3) can be further reduced to 2(mod 3); and you can also think it as this way; 122 = 120 + 2 = 120 -3 +3 +2 = 117 +5 = 3*39 +5 which is also correct.
Sir, 143*4 mod 9 also congruent to 5 mod 9. so why should i have to do 4^-1 in both sides? while i found X then we need to just multiply the X into 4 or the value that is given. Simple.
Sir, I think there is an error in this video. X=143 indeed solves the problem, but X=53 does it also, and it is smaller. I think the issue is in the simplification step @ 5:26. You see, X = 13 solves the original equation 2X ≡ 6 (mod 20), but it's not a solution to X ≡ 3 (mod 20). I believe the 2nd equation simplifies to X ≡ 3 (mod 10) instead.
in the second equation (2x = 6 mod 20) when you divide both side by 2, you must divide 20 by GCD(2,20) so the result would be (x=3 mod 10)
Yes I too felt the same ...Hope he will reply and clarify doubts of thousands @nesoacademy
15 = 5 mod 10 if we common out 5 then it will be 3 = 1 mod 2 and not 3 = 1 mod 10.... so getting answer correct at the end of video is just a coincidence .... in short we are right 4
big disappointment... this channel is usually good. they should re-make this video
You are correct
Answer to H.W. : X = 122
For Verification : 5 mod 3 is reducible to 2 mod 3. So, X = 122 holds for given congruent equations.
In Homework question, please tell me what is the value of M1^-1, M2^-1 and M3^-1?
@@preeti4534 1,2,3 respectively.
BUT 122 IS NOT THE PRIME NUMBER
@@preeti4534 m1 inverse=1,m2inverse=2,m3=25.
HOw can we reduce the 5 mod 3 to 2 mod 3
In the second set equations 4x=9q1+5,2x=20q2+6
x=53 is the minimum value
You explained clearly sir... Keep putting new vedios.. Keep going up.. Continue rocking..
Put more vedios.. Continue
I will always support you sir. ..
Very soon you will reach more & more subscribers... 🎉
All the best...
Clear explanation ❤
Answer 122, M1 = 55, M1^-1 = 1, M2 = 33, M2^-1 = 2, M3 = 15, M2^-1 =3, M = 165
X=122
In equation 2, we need to replace 2x = 6 mod 20 by x = 3 mod 10 for getting correct answer. (please note that 6 cannot be directly divided by 2. there will be change in mod part as well)
5:26 sorry, but x=3 mod 20 is just one of the 2 solutions. The other one is 13 mod 20.
Best teacher 👍
Great teach ❤
@5:38 , the equation written is wrong , it should be x= 3 mod10 , we also have to divide inside the mod, eg. 10=2mod8 implies 5=1mod4 , not 5=1mod8 , so the solution is wrong
It's incomplete, there will be 2 solutions
3mod20 and 13mod20
1:45 yes we can, but what X is? what we are calculating. where those other come? like this is private key and this cipher and x answer is private key. like real example not just this numbers. what they are?. this was exact same video but use diff numbers that other. what have to do network security?
2nd equation should be X=3(mod 10)
Because cancel 2 from mod as well
Final answer is 53
Yes I have the same doubt
Correct.
Yea when I saw the equation just now, I immediately went to the comment section.
5:27 i feel like we can't cancel 2 there. We can only do that if 2 and 20 are relatively prime but they are not.
Yes wrong it will be x=3(mod10)
Yes , you both are correct@@sanjaymandal1698
The given problem states that \( X = 122 \) for mod 3, 5, and 11. We can't get \( a1, a2, a3 \) as 5, 2, and 1, respectively. Therefore, the question is incorrect. The correct values for \( a1, a2, a3 \) should be 2, 2, and 1. hope you understand , if i am wrong than let me know the correction..
my freind, thank you so much, you explain everything pretty simple and understandable...
53 is the smallest positive solution. Increments by 90 will be valid as well
nice explaination sir
😃😃😃thanks sir
thank u so much sir🙏
I went through the problem but fixed the error at 5:25 and got x = 53 (mod 90). Which works, but it would nice for the modulus to be 180. Here's what I did:
x = 53 (mod 90)
x = 90k + 53 for some integer k.
k is either even or odd, in other words leaves a remainder of 0 or 1 when divided by 2.
If k is even:
x = 90k + 53 = 90(2m) + 53 = 180m + 53, where m is another integer. So, x = 53 (mod 180) is a solution.
If k is odd:
x = 90k + 53 = 90(2n + 1) + 53 = 180n + 90 + 53 = 180n + 143, where n is an integer. So, x = 143 (mod 180) is the second solution.
Thank u sir
Nice!
HW: X = 122, but when 122 is divided by 3, we get 2 as the remainder, not 5 Which doesn't fit for x = 5 (mod 3). Or is X = 5 (mod 3) equivalent to X = 2 (mod 3) since 5 / 3 = r2?
its correct. 5(mod 3) can be further reduced to 2(mod 3); and you can also think it as this way; 122 = 120 + 2 = 120 -3 +3 +2 = 117 +5 = 3*39 +5 which is also correct.
@@pathumdilharadissanayake5032 thx, cuz I rechecked my math, didn't see anything wrong, saw 122 in another comment, then was confused by the 5.
I think the correct ans of 2nd last (6:00) quo is 11
Well explained
Answer of homework question:-
X=122 mod 165
please share solution pic
There is no x=122mod165, it's x=122//
THANKU FOR SHARING WRONG INFO 😊😊😊😊
I LOST MY 4 MARKS BEACUSE OF YOU
JUST REALISED U CANT JUST DIVIDE BOTH SIDES BY 2
THANKU FOR YOUR EFFORTS
😂😂
at 3:24 is not clear what you are doing there...
Lovely❤🎉🎉🎉thank sir🎉
Sir, 143*4 mod 9 also congruent to 5 mod 9. so why should i have to do 4^-1 in both sides? while i found X then we need to just multiply the X into 4 or the value that is given. Simple.
The multiplicative inverse of 2 mod 20 does not exist, how did you make the coefficient 2x equal to 1?
thanks
X=122
Home question: Change 1st equation X=1(mod 3) then X will be 67 holds true for all
HW: X=122
I am getting 112
@@vamshireddy3830
112 ≡ 2 (mod 11) since 10 x 11 = 110.
Same answer for me, but 122 mod 3 is 2, not 5.
Edit: Nevermind. 5 mod 3 is 2.
Sir, I think there is an error in this video.
X=143 indeed solves the problem, but X=53 does it also, and it is smaller.
I think the issue is in the simplification step @ 5:26.
You see, X = 13 solves the original equation 2X ≡ 6 (mod 20), but it's not a solution to X ≡ 3 (mod 20).
I believe the 2nd equation simplifies to X ≡ 3 (mod 10) instead.
Yes.
Can I have your phone number please?
2.X=6(mod20) X=13 , a1=8,a2=13,m1=9,m2=20,M=180,M1=20,M2=9 ==> 20.M1i=1(mod9) , M1i=5 and 9.M2i=1 , M2i=9 as equation X=(8.20.5+13.9.9)mod180 , 1853 (mod180)=53 so X=53
Yes this must be correct
1:10 we can read. there no need this1 this2 this thre bladiblabla
X=30
why i am getting M I of 20 mod 9 equal to 9 using extended euclidian theorem instead of 5
Homework answer: 122.
please share solution pic
Please give the explanation of homework question.
h/w ans is 137
please upload next videos also ASAP
I got X=122 for home work sum but in verification for first equation i.e 122 mod 3=2 but in question it is 5.How is it possible??
because 5(mod3)=2(mod3)
5%3 == 2%3
answer is 53
sir what is the meaning of neso ? it is a simple name ?
no it's a SO SIMPLE name
M1 and M1 inverse gets cancelled right
Formula ka Sign Wrong hai 🙂 Congruent sign hoga, Equal sign nahi hoga
Just ignore that brother
uska naam to nhi lunga lekin itna jrur kahunga ki he falani
4X=5mod9, 2X=6mod20
sol=143(i love you)
2x=6mod20 -> x=3mod10 not 3mod20
Hw. Ans. 287
Im not able to find M3 inverse
Home work answer is=122
m3 inverse for homework is 25??
H/W Answer is x = 40 .
Sir ji 😢 mere se m1×m inverse nhi nikal rha h solution btao please 😢
ans 53
122
for hw question im getting 98mod15
X=12
113
Maybe 7?
X=368
bro doesn't know how to pronounce a simple a and x, what the hell is yae yex?
❤🥰🥰❤🥰🥰❤🥰🥰❤
X=77
Sir apne mod toh niklana sikhaya hi nhi
112
This video is wrong
x=122
HW
X =122
X=122
x = 122
X = 110
122
X=122
122
X=122
122
X=122
122