Ad
related to: grade 2 mod 3 lesson 4 problem set 12 1
Search results
Results from the WOW.Com Content Network
For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 / 2 ≡ 2 (mod 3). Equivalently, 2n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 2 (mod 3). Conjecturally, this inverse relation forms a tree except for a 1–2 loop (the inverse of the 1–2 loop of the function f(n) revised as indicated above).
Quadratic Reciprocity (Legendre's statement). If p or q are congruent to 1 modulo 4, then: is solvable if and only if is solvable. If p and q are congruent to 3 modulo 4, then: is solvable if and only if is not solvable. The last is immediately equivalent to the modern form stated in the introduction above.
Let p be an odd prime. The quadratic excess E (p) is the number of quadratic residues on the range (0, p /2) minus the number in the range (p /2, p) (sequence A178153 in the OEIS). For p congruent to 1 mod 4, the excess is zero, since −1 is a quadratic residue and the residues are symmetric under r ↔ p − r.
Adding 4 hours to 9 o'clock gives 1 o'clock, since 13 is congruent to 1 modulo 12. In mathematics, modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when reaching a certain value, called the modulus. The modern approach to modular arithmetic was developed by Carl Friedrich Gauss in his book Disquisitiones ...
On the other hand, the primes 3, 7, 11, 19, 23 and 31 are all congruent to 3 modulo 4, and none of them can be expressed as the sum of two squares. This is the easier part of the theorem, and follows immediately from the observation that all squares are congruent to 0 (if number squared is even) or 1 (if number squared is odd) modulo 4.
However, the linear congruence 4x ≡ 6 (mod 10) has two solutions, namely, x = 4 and x = 9. The gcd(4, 10) = 2 and 2 does not divide 5, but does divide 6. Since gcd(3, 10) = 1, the linear congruence 3x ≡ 1 (mod 10) will have solutions, that is, modular multiplicative inverses of 3 modulo 10 will exist. In fact, 7 satisfies this congruence (i ...
For each of them, compute the remainder by 4 (the second largest modulus) until getting a number congruent to 3 modulo 4. Then one can proceed by adding 20 = 5 × 4 at each step, and computing only the remainders by 3. This gives 4 mod 4 → 0. Continue 4 + 5 = 9 mod 4 →1. Continue 9 + 5 = 14 mod 4 → 2. Continue 14 + 5 = 19 mod 4 → 3.
There are exactly ⌊p/12⌋ supersingular elliptic curves with automorphism groups of order 2. In addition if p≡3 mod 4 there is a supersingular elliptic curve (with j-invariant 1728) whose automorphism group is cyclic or order 4 unless p=3 in which case it has order 12, and if p≡2 mod 3 there is a supersingular elliptic curve (with j ...
Ad
related to: grade 2 mod 3 lesson 4 problem set 12 1