Solved Question 6 Suppose A 6 Mod 10 And B 3 Mod 10 Chegg Suppose a = 6 (mod 10) and b = 3 (mod 10) where a and b are integers. find the integer c with 0. your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on. question: suppose a = 6 (mod 10) and b = 3 (mod 10) where a and b are integers. find the integer c with 0. Suppose that a and b are integers, a ≡ 11 (mod19), and b ≡ 3 (mod19). find the integer c with 0 ≤ c ≤ 18 such that… sure, i can help with that. first, let's understand what the question is asking. the notation "a ≡ 11 (mod19)" means that when you divide 'a' by 19, the remainder is 11.
Solved 17 Suppose A And B Are Integers With A 6 Mod 20 Chegg Find the integer c with 0 sc 9 such that c = 3a (mod 10). a) 2 b)3 c) 6 d) 8. your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on. question: suppose a = 6 (mod 10) and b = 3 (mod 10) where a and b are integers. find the integer c with 0 sc 9 such that c = 3a (mod 10). Since we know a ≡ 11 (mod 19) and b ≡ 3 (mod 19), we substitute these values and simplify the congruence to c^2 ≡ 8 (mod 19). to solve this quadratic congruence, we can use the chinese remainder theorem to find the two square roots of 8 modulo 19, which are 7 and 12. We can substitute our expression for a into this equation: c = 3a (mod 10) c = 3 (10k 6) (mod 10) c = 30k 18 (mod 10) c = 10 (3k 1) 8 (mod 10) so we see that c is 8 more than a multiple of 10, or in other words: c = 8 (mod 10) but we want to find an integer c with 0 < c < 9, so we need to find the residue of 8 (mod 10) in this range. Question 6 suppose a = 6 (mod 10) and b = 3 (mod 10) where a and b are integers. find the integer c with 0 scs 9 such that c = 3a (mod 10). a) 2 b) 3 d) 8 your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on.
Solved Question 11 Suppose A 6 Mod 10 And B 3 Mod 10 Chegg We can substitute our expression for a into this equation: c = 3a (mod 10) c = 3 (10k 6) (mod 10) c = 30k 18 (mod 10) c = 10 (3k 1) 8 (mod 10) so we see that c is 8 more than a multiple of 10, or in other words: c = 8 (mod 10) but we want to find an integer c with 0 < c < 9, so we need to find the residue of 8 (mod 10) in this range. Question 6 suppose a = 6 (mod 10) and b = 3 (mod 10) where a and b are integers. find the integer c with 0 scs 9 such that c = 3a (mod 10). a) 2 b) 3 d) 8 your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on. Our expert help has broken down your problem into an easy to learn solution you can count on. question: suppose a = 6 (mod 10) and b = 3 (mod 10) where a and bare integers. find the integer k such that k = 3a (mod 10). a) 2 b) 3 c) 6 d) 8. here’s the best way to solve it. note : th …. We will show you two methods of finding 6 mod 10 (6 modulo 10). to differentiate our methods, we will call them the "modulo method" and the "modulus method". We can do this by subtracting 10 from 18 until we get a number less than 10: 18 10 = 8 so we have found that 3a is equal to 8 (mod 10).answertherefore, the integer k that satisfies k = 3a (mod 10) is 8. We first need to find the integer with 0 < c < 9 such that =3a (mod 10). since =6 (mod 10) and b = 3 (mod 10), we can see that c = 2 and 3a = 6.therefore, the integer with 0 < c < 9 is 3.
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