2 = 4(mod3) is a true/false question and the = sign after 2 is the one with the 3 bars the equivalence sign. since i don't know the code to put that in.
like if i get something like (2-4)(mod 7) and (2+9)(mod 10) how would i go about solving those?
2 = 4 (mod 3) is false.
Modulo arithmetic really isn't too bad once you get used to the idea. In fact, we often use modulo arithmetic to tell the time; For example, 13 o'clock = 1 o'clock. And, 20:00 = 8pm. This is all modulo 12.
Two number are equal modulo n if they differ by a multiple of n.
Perhaps the best way to write a number modulo n, is to write it as a number less than n but greater than 0.
If you number is lager than n, subtract multiples of n until you reach a number less than n. If you have a negative number, add multiplies of n until you arrive at a number greater than 0 but less than n.
so what is 4 modulo 3? Well, 4 -3 = 1. So 4 = 1 mod 3.
What is -10 modulo 3? Keep adding multiples of 3 until you reach a number greater than 0 but less than 3. -10 = -7 = -4 = -1 = 2 mod (3)
Addition and subtraction are easy, just work out the answer as usual, then add or subtract multiples of 3 until you arrive at a number greater than 0 but less than n.
hmm that seems understandable,
and makes alot more sense now
how about fractions? n/2 = 4(mod7)
so n = 2(mod7)
n=2(mod7) n=9(mod7) n=16(mod7) n=23(mod 7)
would that be the solution?
then it comes regular equations like
3=(n-4)(mod7)
so 3n-12(mod7)
n=-3(mod7) n=5(mod7) n=12(mod7) n=19(mod7)
is my math correct? lol thanks in advance!
For the first one, you get the idea, but you made a small mistake.
If you have
then, as you would do usually, multiply (not divide!) both sides by 2, and you get
Now, as Ant explained, this is not the only answer. In fact, the "best" or most "standard" answer is the one where n is equivalent to some number between 0 and 7; in this case, that would be .
For the second equation, note that
is the same equation as
.
So, we again can just solve it like we would any regular equation, by adding 4 to both sides:
So our answer is (or better: ).