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How is $3\\equiv 3\\bmod{5}$ - Mathematics Stack Exchange Having that in mind 5%2 mean that we successively take 2 objects from a bag containing 5 objects 1st we will take the 2 objects and leaving behind 3 objects, then we take another 2 object leaving behing 1 object in the bag, since we can't take 2 objects any more, SINCE there is only one remaining hence 5%2=1 applying the same analogy and solving for 3%5 we initially have 3 object in a bag
Solve for $x$: $4x = 6~(\\mod 5)$ - Mathematics Stack Exchange Very simple question, but what is the proof that x y mod m == ((x mod m) y) mod m? 3 Determine the smallest multiple of 9 which divided by each of the numbers 2, 5 and 11 leaves a remainder 1
How can I find a mod with negative number? [duplicate] e g $43=-2 (mod\;5)$ You can express as above $43 = 8 \times 5 + 3$, or $43 = 9 \times 5 - 2$ In this last expression you are "short" $2$ to complete the module $5$, but you are "over $3$" to the previous complete module $40$ Hence the positive remainder is $5-2 =3$ (i e Module plus the negative remainder) Operationally would be to use the
number theory - Quadratic residues, mod 5, non-residues mod p . . . Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
How is this a field? - Mathematics Stack Exchange Both operations are commutative and associative To see that every non-zero element has a multiplicative inverse you could just write out the multiplication table The additive inverse is just the negative (as usual) So $-2 = -1*5+3$ and hence $-2 \equiv 3\text{ mod }5$ So the additive inverse of 2 is 3 (because $2+3=5\equiv 0$)
How to read mod operation? (remainders and congruences) $\begingroup$ When you write $7\bmod{(5)}=2\bmod{(5)}$ (you shouldn't use $\:\equiv\:$ in this case), it means that $7$ and $2$ have the same congruence class F y i , the notation $èquiv` is obtained with the command \equiv The notation of your last example is not quite correct for me; one should write mod 5 only once, at the end $\endgroup$
How do you solve the following a - b (mod5) = a + b (mod5)? Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers