Solving congruences, 3 introductory examples

Solving congruences, 3 introductory examples,

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Equipment:
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[rb]

You can solve the third one either way,

1)
4x = 3 (mod 5)
-1x = 3 (mod 5)
x = -3 (mod 5)
x = 2 (mod 5)

2)
4x = 3 (mod 5)
4x = 3 + 5 (mod 5)
4x / 4 = 8 / 4 (mod 5) (gcd(4, 5) = 1)
x = 2 (mod 5)

It's interesting how you can simplify either side by multiples of 5 to get the answer, really enforces the idea that it is "mod 5".

[BigDBrian]

note, for this comment = will replace the congruence symbol
4x = 3 (mod 5)
4x = 8 (mod 5)
x = 2 (mod 5)

[FrogworfKnight]

Not certain if this is a weird method for the second one but...
4x≡2(mod 5)
2x≡1(mod 5) [divide by two, since gcd(2, 5)=1 and 2/2 gives an integer answer]
2x≡6(mod 5) [increasing 1 by one modular 5 cycle to 6]
x≡3(mod 5) [divide by two, since again gcd (2, 5)=1 and 6/2 gives an integer answer]

[elsavelaz]

I wish it had the final solution to self-check understanding, because if it was "so easy, " I wouldn't have googled help. Cute vid, good explanation, just please finish the examples!

[hungryfareasternslav]

3:38 When you finish the number theory

[muhammadhamid]

4x=3 mod 5
(5x-x)=3 mod 5
-x=3 mod 5
x=-3 mod 5
x=2 mod 5

Thank you so much Black Pen Red Pen.

[akshat]

For the third one, just add 5 to the right side and it becomes the same as the first equation.

[rexevan]

So that's where the black pen red pen yay coming from

[jonashammerich]

I really love your mic! still!!! And that intro was cute!!

[deidara_]

Trick: 4 = -1 (mod 5)
4x = 8 (mod 5) <=> -x = 8 (mod 5)
4x = 2 (mod 5) <=> -x = 2 (mod 5)
4x = 3 (mod 5) <=> -x = 3 (mod 5)

Of course all of them are trivial
-x = 8 (mod 5) <=> x = 5k-3 for any integer k (8 = 3 mod 5)
-x = 2 (mod 5) <=> x = 5k-2 for any integer k
-x = 3 (mod 5) <=> x = 5k-3 for any integer k

[diablo]

can u tell how can I solve similar questions where equation is x^2 = a (mod p), here value of x and a is known, how can i find p

[emmanuelmercy]

The one I even want u to do u did not do it self🤨

[Danicker]

For the second one you could also do:
4x = 2mod5
4x = 12mod5 (+10)
x = 3mod5 (/4)

[sugarfrosted]

alternative method would be multiplying by the inverse of 4 aka 4. Granted this is harder in the general case because computing the inverse usually takes more time.

[sergioh]

2 videos on modular arithmetic back after back! :D thanks again. I really like the videos a lot thnx. Mainly cuz I actually like modular arithmetic haha thnx

[chancerc]

Assume the equal signs are congruences, due to keyboard limitations.

Given 4x=2 (mod 5),

Can you just do:

4x=2=12 (mod 5)
=> 12/4=3, hence
{xEZ: x=3 (mod 5/1)} is set of all solutions.

[shandyverdyo]

I'm very grateful to u.
Thanks and nice timing that i need this one.

[cerendemir]

This is just what I needed. Thank you!

[Ethan-mjwy]

Thank you for all the high quality videos bprp, they are much appreciated 💜

[javierlim]

The answer to the last one is 2(mod5) right?