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Find the least number which when divided by 7,9,11 produce 1,2,3 as reminders .

I tried a method which works only when the difference between the elements is same. But in this case 7-1=6 , 9-2=7, 11-3=8 . So that method is not working.

Please, provide the way by which I can solve the problem. Answer of the problem is 344.

learner
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2 Answers2

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Here $\,m_i-x_i\,$ is not constant but, rather $\,m_i - \color{#c00}2x_i = 5,\, $ so $\,\color{#c00}2x_i\equiv -5\pmod{m_i},\,$ so

$\quad 2x \equiv -5\,$ mod $\,7,9,11\iff x \equiv \dfrac{-5}{2}\equiv \dfrac{7\cdot 9\cdot 11-5 }{2}\equiv 344 \pmod{7\cdot 9\cdot 11}$

Bill Dubuque
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  • I don't understand this part 2x=-5 mod . What is it ? – learner Apr 29 '15 at 18:28
  • @learner $\ {\rm mod}\ 5!:\ x\equiv 1\iff 2x\equiv 2\equiv -5,\ $ It is the same idea as you mention, except that to get a constant value we need to $\rm\color{#c00}{doubte}$ the $,x_i,$ before subtracting - see my edit above. – Bill Dubuque Apr 29 '15 at 18:52
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This may be a bit longer, but it is a bit more mechanical of a solution.

To solve $$ \begin{align} x&\equiv1\pmod{7}\\ x&\equiv2\pmod{9}\\ x&\equiv3\pmod{11} \end{align} $$ we can solve three simpler problems. To solve the simpler equations, we can use the Extended Euclidean Algorithm as implemented in this answer.

$$ \begin{align} x_7&\equiv1\pmod{7}\\ x_7&\equiv0\pmod{9}\\ x_7&\equiv0\pmod{11} \end{align} $$ $x\equiv0\pmod{9}$ and $x\equiv0\pmod{11}\iff x\equiv0\pmod{99}$. $$ \begin{array}{r} &&14&7\\\hline 1&0&1&-7\\ 0&1&-14&99\\ 99&7&1&0 \end{array} $$ Therefore, $x_7=99$.

$$ \begin{align} x_9&\equiv0\pmod{7}\\ x_9&\equiv1\pmod{9}\\ x_9&\equiv0\pmod{11} \end{align} $$ $x\equiv0\pmod{7}$ and $x\equiv0\pmod{11}\iff x\equiv0\pmod{77}$. $$ \begin{array}{r} &&8&1&1&4\\\hline 1&0&1&-1&2&-9\\ 0&1&-8&9&-17&77\\ 77&9&5&4&1&0 \end{array} $$ Therefore, $x_9=154$.

$$ \begin{align} x_{11}&\equiv0\pmod{7}\\ x_{11}&\equiv0\pmod{9}\\ x_{11}&\equiv1\pmod{11} \end{align} $$ $x\equiv0\pmod{7}$ and $x\equiv0\pmod{9}\iff x\equiv0\pmod{63}$. $$ \begin{array}{r} &&5&1&2&1&2\\\hline 1&0&1&-1&3&-4&11\\ 0&1&-5&6&-17&23&-63\\ 63&11&8&3&2&1&0 \end{array} $$ Therefore, $x_{11}=-252$.

We can put the three simpler answers together by $$ x\equiv x_7+2x_9+3x_{11}=-349\equiv344\pmod{693} $$ That is $$ \begin{bmatrix} 1\\0\\0 \end{bmatrix} + 2\begin{bmatrix} 0\\1\\0 \end{bmatrix} + 3\begin{bmatrix} 0\\0\\1 \end{bmatrix} = \begin{bmatrix} 1\\2\\3 \end{bmatrix} \begin{array}{l} \pmod{7}\\ \pmod{9}\\ \pmod{11} \end{array} $$

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robjohn
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