Lovely congruence

{ 7 x + 3 y 10 (mod 16) 2 x + 5 y 9 (mod 16) \large \begin{cases} 7x+3y \equiv 10 \text{ (mod 16)} \\ 2x+5y \equiv 9 \text{ (mod 16)} \end{cases}

If x x and y y are the numbers satisfying the system of linear congruence above, find the sum of the remainders when x x and y y are divided by 16 individually.


The answer is 10.

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1 solution

{ 7 x + 3 y 10 (mod 16) . . . ( 1 ) 2 x + 5 y 9 (mod 16) . . . ( 2 ) \begin{cases} 7x + 3y \equiv 10 \text{ (mod 16)} & ...(1) \\ 2x + 5y \equiv 9 \text{ (mod 16)} & ...(2) \end{cases}

{ ( 1 ) × 5 : 35 x + 15 y 50 (mod 16) . . . ( 3 ) ( 2 ) × 3 : 6 x + 15 y 27 (mod 16) . . . ( 4 ) \begin{cases} (1) \times 5: & 35x + 15y \equiv 50 \text{ (mod 16)} & ...(3) \\ (2) \times 3: & 6x + 15y \equiv 27 \text{ (mod 16)} & ...(4) \end{cases}

( 3 ) ( 4 ) : 29 x 23 (mod 16) 13 x 7 (mod 16) x 3 \begin{aligned} (3)-(4): \quad 29x & \equiv 23 \text{ (mod 16)} \\ 13x & \equiv 7 \text{ (mod 16)} \\ \implies x & \equiv 3 \end{aligned}

( 1 ) : 21 + 3 y 10 (mod 16) 5 + 3 y 10 (mod 16) 3 y 5 (mod 16) y 7 \begin{aligned} (1): \quad 21 + 3y & \equiv 10 \text{ (mod 16)} \\ 5 + 3y & \equiv 10 \text{ (mod 16)} \\ 3y & \equiv 5 \text{ (mod 16)} \\ \implies y & \equiv 7 \end{aligned}

x + y 3 + 7 10 (mod 16) \implies x + y \equiv 3 + 7 \equiv \boxed{10} \text{ (mod 16)}

x is congruent to 3 and y is congruent to 7. Could you please change the 'equal to' to reflect the congruence.

Siva Bathula - 4 years, 4 months ago

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Thanks. I have changed it. Are x 3 x \equiv 3 and y 7 y \equiv 7 correct? I am new to this.

Chew-Seong Cheong - 4 years, 4 months ago

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Yes sir, I wouldn't dare to find mistake in your solutions! :P

Siva Bathula - 4 years, 3 months ago

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