![Modular (Remainder) Arithmetic n = qk + r (for some k; r < k) eg 37 = (2)(17) + 3 Divisibility notation: 17 | n mod k = r 37 mod 17 = ppt download Modular (Remainder) Arithmetic n = qk + r (for some k; r < k) eg 37 = (2)(17) + 3 Divisibility notation: 17 | n mod k = r 37 mod 17 = ppt download](https://images.slideplayer.com/26/8780672/slides/slide_11.jpg)
Modular (Remainder) Arithmetic n = qk + r (for some k; r < k) eg 37 = (2)(17) + 3 Divisibility notation: 17 | n mod k = r 37 mod 17 = ppt download
![discrete mathematics - Why doesn't the author subtract everything by two first before applying modulus? - Mathematics Stack Exchange discrete mathematics - Why doesn't the author subtract everything by two first before applying modulus? - Mathematics Stack Exchange](https://i.stack.imgur.com/NpjFa.png)
discrete mathematics - Why doesn't the author subtract everything by two first before applying modulus? - Mathematics Stack Exchange
![SOLVED: The Chinese Remainder Theorem 4 Example: Consider the 3 congruences from Sun-Tsu's problem: X=2 mod 3), x=3 mod 5),X=2 mod 7). Let m =3. 5 . 7 105, M1 m/3 = SOLVED: The Chinese Remainder Theorem 4 Example: Consider the 3 congruences from Sun-Tsu's problem: X=2 mod 3), x=3 mod 5),X=2 mod 7). Let m =3. 5 . 7 105, M1 m/3 =](https://cdn.numerade.com/ask_previews/f3f912f0-a507-4168-a4b2-01104dfac95c_large.jpg)
SOLVED: The Chinese Remainder Theorem 4 Example: Consider the 3 congruences from Sun-Tsu's problem: X=2 mod 3), x=3 mod 5),X=2 mod 7). Let m =3. 5 . 7 105, M1 m/3 =
![elementary number theory - How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? - Mathematics Stack Exchange elementary number theory - How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? - Mathematics Stack Exchange](https://i.stack.imgur.com/mRBf7.jpg)