3.4: The Chinese Remainder Theorem
- Page ID
- 8834
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, we discuss the solution of a system of congruences having different moduli. An example of this kind of systems is the following; find a number that leaves a remainder of 1 when divided by 2, a remainder of 2 when divided by three and a remainder of 3 when divided by 5. This kind of question can be translated into the language of congruences. As a result, in this chapter, we present a systematic way of solving this system of congruences.
The system of congruences
\[\begin{aligned} && x\equiv b_1(mod \ n_1),\\&&x\equiv b_2(mod \ n_2),\\&&.\\&&.\\&&.\\&& x\equiv b_t(mod \ n_t),\end{aligned}\]
has a unique solution modulo \(N=n_1n_2...n_t\) if \(n_1,n_2,...,n_t\) are pairwise relatively prime positive integers.
Let \(N_k=N/n_k\). Since \((n_i,n_j)=1\) for all \(i\neq j\), then \((N_k,n_k)=1\). Hence by Theorem 26 , we can find an inverse \(y_k\) of \(N_k\) modulo \(n_k\) such that \(N_ky_k\equiv 1(mod \ n_k)\). Consider now
\[x=\sum_{i=1}^tb_iN_iy_i\]
Since \[N_j\equiv 0(mod \ n_k) \ \ \mbox{for all} \ \ j\neq k,\] thus we see that \[x\equiv b_kN_ky_k(mod \ n_k).\] Also notice that \(N_ky_k\equiv 1(mod \ n_k)\). Hence \(x\) is a solution to the system of t congruences. We have to show now that any two solutions are congruent modulo \(N\). Suppose now that you have two solutions \(x_0,x_1\) to the system of congruences. Then \[x_0\equiv x_1(mod \ n_k)\] for all \(1\leq k\leq t\). Thus by Theorem 23, we see that \[x_0\equiv x_1(mod \ N).\] Thus the solution of the system is unique modulo \(N\).
We now present an example that will show how the Chinese remainder theorem is used to determine the solution of a given system of congruences.
Example \(\PageIndex{1}\):
Solve the system \[\begin{aligned} && x\equiv 1(mod \ 2)\\&& x\equiv 2(mod \ 3)\\&& x\equiv 3(mod \ 5).\end{aligned}\] We have \(N=2.3.5=30\). Also \[N_1=30/2=15, N_2=30/3=10 \mbox{and} \ \ N_3=30/5=6.\] So we have to solve now \(15y_1\equiv 1(mod \ 2)\). Thus \[y_1\equiv 1(mod \ 2).\] In the same way, we find that \[y_2\equiv 1(mod \ 3) \mbox{and} \ \ y_3\equiv 1(mod \ 5).\] As a result, we get \[x\equiv 1.15.1+2.10.1+3.6.1\equiv 53\equiv 23 (mod \ 30).\]
Exercises
- Find an integer that leaves a remainder of 2 when divided by either 3 or 5, but that is divisible by 4.
- Find all integers that leave a remainder of 4 when divided by 11 and leaves a remainder of 3 when divided by 17.
- Find all integers that leave a remainder of 1 when divided by 2, a remainder of 2 when divided by 3 and a remainder of 3 when divided by 5.
Contributors and Attributions
Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution (CC BY) license.