SOLUTION: Modular arithmetic ... Find all possible replacements for n for which each congruence is true. a. 6 + n &#8801; 1 (mod 7) b. 3

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Question 271955: Modular arithmetic ...
Find all possible replacements for n for which each congruence is true.
a. 6 + n ≡ 1 (mod 7) b. 3 - n ≡ 4 (mod 7)

c. 2n ≡ 1 (mod 3) d. n/2 ≡ 2 (mod 3) (n/s is a fraction)

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Modular arithmetic ...
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Comment: The fact to remember is (mod n) means n is zero.
So 1 (mod 5) = 6 = 11 = 16 = -4 = - 9
5 in all these numbers is zero so 16 is just 1+3*zero = 1
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Find all possible replacements for n for which each congruence is true.
a. 6 + n ≡ 1 (mod 7)
7 is zero so add it to the 1 to get:
6 + n = 8 (7)
n = 2+7k where k is any integer
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b. 3 - n = 4 (mod 7)
Regular arithmetic applies so solve for "n":
n = 3-4 = -1 (mod 7)
n = -1+7 = 6 (mod 7)
n = 6 + 7k where k is any integer
c. 2n ≡ 1 (mod 3)
3 is zero so add it to the 1 to get:
2n = 4 (mod 3)
Use regular arithmetic to get:
n = 2 + 3k where k is any integer
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d. n/2 ≡ 2 (mod 3) (n/s is a fraction)
Use regular arithmetic to get:
n = 4 (mod 3
n = 1 + 3k where k is any integer
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Cheers,
Stan H.

SOLUTION: Modular arithmetic ... Find all possible replacements for n for which each congruence is true. a. 6 + n &#8801; 1 (mod 7) b. 3 - n &#8801; 4 (mod 7) c. 2n &#8801;

SOLUTION: Modular arithmetic ... Find all possible replacements for n for which each congruence is true. a. 6 + n ≡ 1 (mod 7) b. 3 - n ≡ 4 (mod 7) c. 2n ≡