## View Poll Results: Correct Answer for system of congruence Equations

Voters
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• x = 10 (mod 12)

0 0%
• x = 16 (mod 33)

0 0%
• x = 38 (mod 44)

0 0%
• x = 82 (mod 132)

0 0%

# Thread: Solve this system of congruence equations (POLL)

1. ## Solve this system of congruence equations (POLL)

Hi again,

I got 4 various answers, of which one I was told by my professor has to be used to complete the problem (x = 10 (mod 12)).

Question:
Solve this system of congruence equations:
x congruent to 1 (mod 3)
x congruent to 2 (mod 4)
x congruent to 5 (mod 11).

I set the first two equations equal to each other and got x = 10 (mod 12). My professor told me to use the transitive property and use this answer with x congruent to 5 (mod 11) to find out the final solution.

Which do you believe is the correct answer? (Please comment if you believe the correct answer is not present).

x = 10 (mod 12)
x = 16 (mod 33)
x = 38 (mod 44)
x = 82 (mod 132)

2. ## Re: Solve this system of congruence equations (POLL)

Why is this a poll?

-Dan

3. ## Re: Solve this system of congruence equations (POLL)

Maths is not a matter of opinion, it's a matter of fact.

\displaystyle \left. \begin{aligned}x&=1 \pmod{3} &\implies x &\in \{1, 4, 7, 10\} \pmod{12} \\ x&=2 \pmod{4} &\implies x &\in \{2, 6, 10\} \pmod{12}\end{aligned} \right\} \implies x &= 10 \pmod{12}

Continue.

4. ## Re: Solve this system of congruence equations (POLL)

My method: From x= 5 (mod 11), x= 5+ 11i for any integer i.

Putting that into x= 2 (mod 4), 5+ 11i= 5+ 3i= 2 (mod 4). So 3x= 2- 5= -3= 1 (mod 4) and i= 3 (mod 4). That is, i= 3+ 4k for any integer j. Then x= 5+ 11i= 5+ 11(3+ 4k)= 5+ 33+ 44k= 38+ 44k.

Putting that into x= 1 (mod 3), 38+ 44k= 2+ 2k= 1 (mod 3). So 2k= 1- 2= -1= 2 (mod 3) and k= 1 (mod 3). That is, k= 1+ 3n.
Then x= 38+ 44k= 38+ 44(1+ 3n)= 82+ 132n.

Actually, since this is a "multiple guess" question, you don't need to actually solve the equations. The "moduli", 3, 4, and 11 are all relatively prime so the answer must be some number "modulo" 3*4*11= 132. The only answer that fits that is the last one.

5. ## Re: Solve this system of congruence equations (POLL)

My method: From x= 5 (mod 11), x= 5+ 11i for any integer I.

Putting that into x= 2 (mod 4), 5+ 11i= 5+ 3i= 2 (mod 4). So 3x= 2- 5= -3= 1 (mod 4) and i= 3 (mod 4). That is, i= 3+ 4k for any integer k. Then x= 5+ 11i= 5+ 11(3+ 4k)= 5+ 33+ 44k= 38+ 44k.

Putting that into x= 1 (mod 3), 38+ 44k= 2+ 2k= 1 (mod 3). So 2k= 1- 2= -1= 2 (mod 3) and k= 1 (mod 3). That is, k= 1+ 3n.
Then x= 38+ 44k= 38+ 44(1+ 3n)= 82+ 132n.

Actually, since this is a "multiple guess" question, you don't need to actually solve the equations. The "moduli", 3, 4, and 11, are all relatively prime so the answer must be some number "modulo" 3*4*11= 132. The only answer that fits that is the last one.

And, of course, one way to solve any "multiple guess" is to check each of the suggested answers:
x= 10 (mod 12) is the same as 10+ 12n. If x= 10+ 12n then x= 10+ n= 10 (mod 11), not 5.
x= 16 (mod 33) is the same as 16+ 33n. mod 11, that is 5 but mod 4, it is 0, not 2.
x= 39 (mod 44) is the same as 39+ 44n. mod 11, that is 6, not 5.

x= 82 (mod 132) is the same as 82+ 132n. mod 11 that is 5, mod 4 it is 2, and mod 3, it is 1.