Substitute x in third congruence 30u 26 3 mod 7 Solution is u 6 mod 7 Thus u

Substitute x in third congruence 30u 26 3 mod 7

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Substitute x in third congruence: 30u + 26 3 (mod 7) Solution is u 6 (mod 7) Thus u = 7v+6 [2] From [1] and [2] x = 30(7v+6) + 26 = 210u + 206 Therefore x 206 (mod 210) Note: m = 5*6*7 = 210, so we’re looking for a solution modulo 210
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Large integers and pseudoprimes
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Representing large integers Choosing m 1 ,m 2 , …, m n to be pairwise relatively prime, we can represent any integer 0 < a < m = m 1 m 2 …m n by an n-tuple (a mod m 1 , a mod m 2 , …, a mod m n ) Example: 99, 98, 97, 95 are pairwise relatively prime m = 99 * 98 * 97 * 95 = 89,403,930 Every integer 0…m can be represented uniquely by four remainders 123,684 = (33, 8, 9, 89) 123,684 mod 99 = 33 123,684 mod 98 = 8 123,684 mod 97 = 9 123,687 mod 95 = 89
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Arithmetic with large integers Arithmetic operations are carried out upon such tuples on a component basis, then we need to solve a system of linear congruences 123,684 = (33,8,9,89) 413,456 = (32,92,42,16) 123,684 + 413,456 = (33,8,9,89) + (32,92,42,16) = (65 mod 99, 100 mod 98, 51 mod 97, 105 mod 95) = (65, 2, 51, 10) To compute the sum, we need to solve a system of congruences x 65 (mod 99) x 2 (mod 98) x 51 (mod 97) x 10 (mod 95)
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Fermat’s Little Theorem If p is prime and a is an integer not multiple of p then a p-1 1 (mod p) And for every integer a we have a p a (mod p) Use case: Find 7 222 mod 11 Fermat’s little theorem says 7 10 1 (mod 11) 7 222 = 7 10x22+2 = (7 10 ) 22 x 7 2 = 1 22 x 49 5 (mod 11) Find 6 138 mod 23 Fermat’s little theorem says 6 22 1 (mod 23) 6 138 = 6 22x6+6 = (6 22 ) 6 x 6 6 = 1 6 x 46656 12 (mod 23)
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Fermat to test for prime-ness Fermat’s Little Theorem: 2 n-1 1 (mod n) if n is prime However, in rare cases, the same congruence holds for composite numbers too, e.g., for 341 = 11 x 31 [Question: does that mean that Fermat’s Little Theorem is wrong?] How often does the congruence hold for composite numbers? From all numbers less than 1010
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Pseudorandom numbers Most pseudo-random number generators implemented in computers use the linear congruential method. They generate numbers based on a recurrence: xn+1= (axn+ c) mod m where x0= seed, a = multiplier, c = increment, and m = modulus We can divide xnto m to get a random number in (0,1) Question: how do we know that the mapping works well (does not map different numbers to the same value, and doesn’t go outside the interval)? The values a = 75, m = 231-1 and c = 0 (aka purely multiplicative generator) are used in practice. This generates 231-2 numbers before the first repetition.
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