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Find natural number $x$ so that $$\begin{align}x&\equiv 9\pmod{10}\\ x&\equiv8\pmod9\\ &\ \ \vdots\\ x&\equiv 1\pmod2\end{align}$$

Bill Dubuque
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user119081
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3 Answers3

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\begin{align} x &\equiv 9 \pmod{10} \\ x &\equiv 8 \pmod 9 \\ x &\equiv 7 \pmod 8 \\ x &\equiv 6 \pmod 7 \\ x &\equiv 5 \pmod 6 \\ x &\equiv 4 \pmod 5 \\ x &\equiv 3 \pmod 4 \\ x &\equiv 2 \pmod 3 \\ x &\equiv 1 \pmod 2 \\ \end{align}

Is equivalent to

\begin{align} x &\equiv -1 \pmod{10} \\ x &\equiv -1 \pmod 9 \\ x &\equiv -1 \pmod 8 \\ x &\equiv -1 \pmod 7 \\ x &\equiv -1 \pmod 6 \\ x &\equiv -1 \pmod 5 \\ x &\equiv -1 \pmod 4 \\ x &\equiv -1 \pmod 3 \\ x &\equiv -1 \pmod 2 \\ \end{align}

which is equivalent to

$$x \equiv -1 \mod{\operatorname{lcm}\{2,3,4,5,6,7,8,9,10\}}$$

$$x \equiv -1 \pmod{2520}$$

$$x \equiv 2519 \pmod{2520}$$

Steven Alexis Gregory
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Hint: The unnatural number $-1$ works.

André Nicolas
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  • I ask for natural number not integer – user119081 Mar 16 '14 at 20:44
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    @user119081: Yes, you did. But André's hint is still a good one. How could you find a second integer solution from that one? Could you make the second solution natural? – Charles Mar 16 '14 at 20:45
  • Yes, I know. But from my solution $-1$, which is not a natural number, one can find a natural number solution, indeed all natural number solutions. I am leaving the discovery of that, at least for a while, to you. – André Nicolas Mar 16 '14 at 20:46
  • I believe this should be exist a natural solution by extended Euclidean algorithm – user119081 Mar 16 '14 at 20:48
  • You could use the Chinese Remainder Theorem, but only after dealing with the fact not all your moduli are relatively prime. But that is very much the hard way for these very special numbers. – André Nicolas Mar 16 '14 at 20:50
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    While waiting, perhaps you can think of this. What number(s) can you add to my solution to get a natural number solution? – André Nicolas Mar 16 '14 at 20:52
  • In fact, I saw this in the a school problem book as a hard question. But I could not respond to this question – user119081 Mar 16 '14 at 20:54
  • Let $b$ be the least common multiple of the numbers $2$ to $10$. The positive integer solutions are $-1+bk$, where $k$ ranges over the positive integers. So the smallest positive solution is $2520-1$. An easier positive solution is $10!-1$. – André Nicolas Mar 16 '14 at 20:59
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Since $\,m_i-1\equiv \color{#c00}{-1}\pmod{\!m_i}\,$ we can apply $ $ CCRT = $\rm\color{#c00}{constant}$ case optimization of CRT

$$\begin{align} x\equiv \color{#c00}{-1}\!\!\pmod{\!m_i}&\iff x\equiv -1\!\!\pmod{{\rm lcm}\{m_i\}}\\[.4em] \text{or, without using $\rm{\small CRT\!:}$}\ \ \ {\rm all}\ \ m_i \mid x+1 &\iff {\rm lcm}\{m_i\}\mid x+1 \end{align}\qquad\qquad$$ The latter equivalence is by the Universal Property of LCM (= definition of LCM in general)

Remark $ $ more generally this idea works for linearly related values & moduli: $ $ if $\,(a,b) = 1\,$ then

$$\left\{\,x\equiv d\!-\!ck\!\!\!\pmod{b\!-\!ak}\,\right\}_{k=0}^{n}\!\!\iff\! x\equiv \dfrac{ad\!-\!bc}a\!\!\!\pmod{{\rm lcm}\{b\!-\!ak\}_{k=0}^n}\quad \ $$

$ $ e.g. here $\,\ \underbrace{\left\{\,x \equiv 3-k\pmod{7-k}\,\right\}_{k=0}^2}_{\textstyle{x\equiv 3,2,1\pmod{\!7,6,5}}}\!\!\iff\! x\equiv \dfrac{1(3)-7(1)}1\equiv -4\pmod{210}$

Bill Dubuque
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  • It suffices to find any common multiple $,m,$ of all the moduli $,m_i$ (e.g. their product) and let $,x = m+1,,$ since then $,x-1 = m,$ is divisible by all $,m_i,,$ so, by above, is a solution. – Bill Dubuque Mar 16 '14 at 21:11