Suppose that a and b are integers, a≡11(bmod19), and b≡3(bmod19). Find the integer c with 0≤c≤18 such that c≡2a2+3b2(bmod19)

DefinitionsDivision algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with 0≤r&lt;d such that a=dq+rq is called the quotient and r is called the remainderq=a ÷ dr=abmoddTheorem 5 Let m be a positive integer. If a≡b(bmodm) and c≡d(bmodm), then a+c≡b+d(bmodm) and ac≡bd(bmodm).Solutiona=11(bmod19)b≡3(bmod19)0≤c≤18Use theorem 5:c≡2a2+3b2(bmod19)=2⋅112+3⋅32(bmod19)=2⋅121+3⋅9(bmod19)=242+27(bmod19)=269(bmod19)=3(bmod19)We then obtain c=3 with 0≤c≤18.

We have an answer to "Suppose that a and b are integers, a≡11(bmod19),..."

Rosemary Burns

Open question

2022-08-19

Suppose that a and b are integers,

a \equiv 11 (b mod 19), and b \equiv 3 (b mod 19)

. Find the integer c with

0 \leq c \leq 18

such that

c \equiv 2 a^{2} + 3 b^{2} (b mod 19)

Answer & Explanation

Jaylen Hickman

Beginner2022-08-20Added 7 answers

Definitions
Division algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with

0 \leq r < d

such that a=dq+r
q is called the quotient and r is called the remainder

q = a \div d

r = a b mod d

Theorem 5 Let m be a positive integer. If

a \equiv b (b mod m) and c \equiv d (b mod m), t h e n a + c \equiv b + d (b mod m) and a c \equiv b d (b mod m)

.
Solution

a = 11 (b mod 19)

b \equiv 3 (b mod 19)

0 \leq c \leq 18

Use theorem 5:

c \equiv 2 a^{2} + 3 b^{2} (b mod 19)

= 2 \cdot 11^{2} + 3 \cdot 3^{2} (b mod 19)

= 2 \cdot 121 + 3 \cdot 9 (b mod 19)

= 242 + 27 (b mod 19)

= 269 (b mod 19)

= 3 (b mod 19)

We then obtain c=3 with

0 \leq c \leq 18

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Didn't find what you were looking for?