Suppose that a and b are integers, a ≡ 11(mod 19), and b ≡ 3(mod 19). Find the integer c with 0 <= c <= 18 such that c ≡ 13a(mod 19)

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Definitions

Division algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with 0 <= r < d such that a=dq+r

q is called the quotient and r is called the remainder

q=a div d

r=a mod d

Theorem 5 Let m be a positive integer. If a ≡ b(mod m) and c ≡ d(mod m), then a+c ≡ b+d(mod m) and ac ≡ bd(mod m).

Solution

a=11(mod 19)

b ≡ 3(mod 19)

0 <= c <= 18

Use theorem 5:

c ≡ 13a(mod 19)

=13*11(mod 19)

=143(mod 19)

=10(mod 19)

We then obtain c=10 with 0 <= c <= 18.

Division algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with 0 <= r < d such that a=dq+r

q is called the quotient and r is called the remainder

q=a div d

r=a mod d

Theorem 5 Let m be a positive integer. If a ≡ b(mod m) and c ≡ d(mod m), then a+c ≡ b+d(mod m) and ac ≡ bd(mod m).

Solution

a=11(mod 19)

b ≡ 3(mod 19)

0 <= c <= 18

Use theorem 5:

c ≡ 13a(mod 19)

=13*11(mod 19)

=143(mod 19)

=10(mod 19)

We then obtain c=10 with 0 <= c <= 18.