Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(a) $x≡5\pmod 7\;\;,\; x≡7\pmod{11}\;\;,\;\;x≡3\pmod{13}$

(b) $x≡3\pmod{10}\;\;,\;\; x≡8\pmod{15}\;\;,\;\;x≡5\pmod{84}$


for (a) I have a rough idea how to do it, its like:

$n_1=7,n_2=11,n_3=13$ then $n=7·11·13=1001$

${1001\over7}·k_1≡1mod(7)$

${1001\over11}·k_2≡1mod(11)$

${1001\over13}·k_3≡1mod(13)$

Hence, $k_1=[5], k_2=[4], k_3=[12]$

$x_0=(11·13)·5·5+(7·13)·4·7+(7·11)·(-1)·3=287$

the solution set is $x=x_0+k_n, k\leftarrow\mathbb{Z}, x=887+k(1001)$

Im not sure if im correct, but i just follow the standard procedure

(b) for this one, what should I do the the $n$'s are not coprime, or prime???

Thank you!!

share|improve this question
add comment

3 Answers

up vote 2 down vote accepted

You've got the idea...

Apply prime factorization to each of the moduli $n$, omit common factors, proceed in much the way you did for (a), and transform each statement in the system into equivalent congruences for the prime powers: (i.e. to a prime residue system). E.g. $$x \equiv 3 \pmod{10} \iff x \equiv 3 \pmod 2 \;\text{and}\; x \equiv 3 \pmod 5\,.$$

And of course, you'll then want to use the Chinese Remainder Theorem and/or the extended CRT.

share|improve this answer
    
Presumably the final $\iff$ should be "and". –  Math Gems Feb 4 '13 at 6:32
add comment

Reduce all the statements to equivalent statements to a prime residue system And then use the chinese remainder theorem

share|improve this answer
    
finding prime factorization of each $n$ then omit the common one? –  Paul Feb 4 '13 at 1:51
    
Thats how id do it. factorize and the make the statement into equivalent statements for the prime powers. For example: $x\equiv 3mod 14$ if and only if $x\equiv 3 mod 7$ and $x \equiv 3 mod 2$ –  user4140 Feb 4 '13 at 2:00
add comment

Case (b) you'd apply the extended chinese remainder theorem. In this case, as $3 \equiv 8 \pmod{\gcd(10, 15)}$, there is a solution, and it will be unique modulo $\mathrm{lcd}(10, 15, 84)$.

Essentially you solve the system $x \equiv 3 \pmod{15}$ and $x \equiv 5 \pmod{84}$, and patch up the solution to agree with your equivalence modulo 10.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.