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Recently I encountered a sum which is as follows:

Let a number when divided by 9 gives remainder 1 , when divided by 11 gives remainder 2 and when divided by 13 gives remainder 3. so Find ot that number ?

Is there a general method to solve this sum ? Also if we want to generalize this sum by placing three numbers with 9,11 and 13. How can we do it ? (I think it might be broad question) But How to generalize it ? Hints or hints toward a particular method will be appreciated. Thanks.
My attempt:
I have a quite idea that it might be able to solve using chinese remainder theorem. But I am not able to successfully use it.

Bill Dubuque
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Murtuza Vadharia
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3 Answers3

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Your system is has congruences $x \equiv a_i \pmod m_i$:

\begin{align} x & \equiv 1 \pmod{9} \\ x & \equiv 2 \pmod{11} \\ x & \equiv 3 \pmod{13} \end{align}

We apply the Chinese remainder theorem (link):

Then $\DeclareMathOperator{lcm}{lcm}M = \lcm(9,11,13) = 1287$.

$M_1 = M/m_1 = 1287/9=143$, $M_2 = M/m_2 = 1287/11 = 117$, $M_3 = M/m_3 = 1287/13 = 99$.

Then we use the extended Euclidean algorithm to solve $$ s_i m_i + t_i M_i = 1 $$ which gives $(s_1, t_1) = (16,-1)$, $(s_2, t_2) = (32,-3)$ and $(s_3, t_3) = (-38,5)$.

For $e_i = t_i M_i$ we then get $e_1 = -143$, $e_2 = -351$, $e_3 = 495$.

A solution is $$ x = \sum_i a_i e_i = 640 $$ All solutions are $$ x = 640 + k \, 1287 \quad (k \in \mathbb{Z}) $$

mvw
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Follows immediately by CCRT for congruences in A.P. (Arithmetic Progression), i.e.

$$\begin{align}x \equiv 1\!+\!j\!\!\!\pmod{\color{#c00}{9\!+\!2j}}\! \iff &\, 2x \equiv 2\!+\!\color{#c00}{2j} \equiv 2\!\color{#c00}{-\!9} \equiv -7\!\!\!\pmod{\!1287 \!=\! 9(11)13}\\[.3em] \iff &\, x \equiv \dfrac{-7}2\equiv\dfrac{1287\!-\!7}2\equiv 640 \end{align}$$

Bill Dubuque
  • 272,048
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The problem has the form

$$x\equiv \begin{cases} a\mod A\\ b\mod B\\ c\mod C\\ \end{cases}$$

so, the Chinese remainder theorem seems clearly called for. However, in this case, the numbers $a,b,c$ are in arithmetic progression, as are the numbers $A,B,C$. That makes it possible to simplify things a bit before doing any explicit arithmetic with actual numbers.

Writing $a=b-d$ and $c=b+d$ for one progression and $A=B-D$ and $C=B+D$ for the other we have

$$u=x-b\equiv \begin{cases} -d\mod(B-D)\\ 0\mod B\\ d\mod(B+D) \end{cases}$$

The central congruence means we have $u=kB$ for some $k$. Noting that we can write $kB$ as $k(B-D)+kD$ and $k(B+D)-kD$, we find that there is a single congruence to solve:

$$kD\equiv-d\mod(B^2-D^2)$$

(To be more precise, the modulus is lcm$(B-D,B+D)$, but in the OP's problem $D=2$ and $B$ is odd.) Applied to the OP's problem we have

$$2k\equiv-1\mod117$$

which has $k=58$ as one obvious solution, from which we find $u=11k=638$ and thus $x=u+2=640$.

Barry Cipra
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