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anonymous
 one year ago
given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true?
anonymous
 one year ago
given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true?

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Below is not required but it helps in relating congruences and divisibility \[x\equiv 1 \pmod 3 \implies 3 \mid (x1) \implies 3\mid [(x1)  3] \\\implies 3\mid (x4)\implies x\equiv 4\pmod{3}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1modulo 3 looks like [0,1,2] though?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1like on x ≡ 1 (mod 3). it's x is congruent to the remainder 1 in modulo 3. I can see that part... it's x ≡ 4 (mod 3) that's giving me a hard time... because that's x is congruent to the remainder 4 in modulo 3.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Notice that \(3\mid (3k)\). This is same as saying \(3k\equiv 0 \pmod{3} \). Therefore \(x\equiv x+0\equiv x+3k\pmod{3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so basically I can add multiple of 3 on either sides, and it still holds?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like x + 3 ≡ 1 (mod 3) holds?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, congruence is just a neat notation that hides the messy divisibility rules

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(x\equiv y\pmod{n}\) and \(y\equiv z\pmod{n}\) implies \(x\equiv z\pmod{n}\) transitive property holds, so you can replace \(3\) by \(0\) in modulo 3

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Whenever I use mods I end up just subtracting off the number until I have some number less than what I want, like this: \[9 \equiv 5 \equiv 1 \mod 4 \] So here I just kept subtracting 4 off, rather than trying to think "ok what's the remainder if I divide by 4?" There are a bunch of cute tricks you can use too, like: \[100^{2523672} \mod 3\] just rewrite 100 as 3*33+1 and we have: \[(3*33+1)^{klfaljag} \mod 3\] \[(0*33+1)^{lsdjf0aug} \mod 3\] \[1 \mod 3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. I know it seems kinda trivial but I just want to make sure I really understand it. x + 3k ≡ 1 + 3s (mod 3), for some integer k,s, holds ? right? ^^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually for *all* integer s,k

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3x + 3k ≡ 1 + 3s (mod 3) thats not so important, all you need to do is to refer back to the definition : \[a\equiv b\pmod{n}\iff n\mid (ab)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty i'm not quite there yet but I might find that useful one day ^^

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3the moment you see 3k, and if you're in modulo 3, you see that 3k is divisible by 3 : \[3\mid (3k) \iff 3k\equiv 0 \pmod{3}\] so you replace \(3k\) by \(0\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1A way to help think of it is all even and all odd numbers can be represented as 2n or 2n+1. So you can see in mod 2, all even numbers are 0 and all odd numbers are 1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 ah i see. x ≡ 4 (mod 3) x ≡ 1 + 3 (mod 3) and since 3 is a multiple of 3, replace 3 with 0. x  99 ≡ 1 (mod 3). And since 99 is also a multiple of 3, replace it with 0. and x + 3k ≡ 1 + 3s (mod 3), replace 3k and 3s with 0.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, but careful, congruence relations don't work in both directions always

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3``` given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true? ``` you should start with whats given, not what you want to prove

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah yes. I just need to see what you can validly add to either side (or both). So in general, a ≡ b (mod c) implies a ≡ b + kc (mod c) implies a + kc ≡ b (mod c) implies a + kc ≡ b + sc (mod c)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yep, adding/subtracting a multiple of "c" to a side(s) wont change the congruence because : \[c\mid (ab) \implies c\mid [(ab) + kc]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep! I understand everything now. Thank you!
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