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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 ...
For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f 2 (4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting ...
In additive number theory, Fermat 's theorem on sums of two squares states that an odd prime p can be expressed as: with x and y integers, if and only if. The prime numbers for which this is true are called Pythagorean primes . For example, the primes 5, 13, 17, 29, 37 and 41 are all congruent to 1 modulo 4, and they can be expressed as sums of ...
The simplest fraction 3 / y with a three-term expansion is 3 / 7 . A fraction 4 / y requires four terms in its greedy expansion if and only if y ≡ 1 or 17 (mod 24), for then the numerator −y mod x of the remaining fraction is 3 and the denominator is 1 (mod 6). The simplest fraction 4 / y with a four-term ...
[1] For example, the expression "5 mod 2" evaluates to 1, because 5 divided by 2 has a quotient of 2 and a remainder of 1, while "9 mod 3" would evaluate to 0, because 9 divided by 3 has a quotient of 3 and a remainder of 0. Although typically performed with a and n both being integers, many computing systems now allow other types of numeric ...
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.
Note that a d ≡ 1 (mod p) holds trivially for a ≡ 1 (mod p), because the congruence relation is compatible with exponentiation. And a d = a 2 0 d ≡ −1 (mod p) holds trivially for a ≡ −1 (mod p) since d is odd, for the same reason. That is why one usually chooses a random a in the interval 1 < a < p − 1.
The greedy algorithm for Egyptian fractions finds a solution in three or fewer terms whenever is not 1 or 17 mod 24, and the 17 mod 24 case is covered by the 2 mod 3 relation, so the only values of for which these two methods do not find expansions in three or fewer terms are those congruent to 1 mod 24. [12]