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I'm having trouble understanding the steps involved to do this question so any step by step reasoning in solving the solution would help me study for my exam.

Thanks so much!

$$x\equiv 6 \pmod{14}$$

$$x\equiv 24 \pmod{29}$$

Michael
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  • Have you covered the Chinese Remainder Theorem? – Jyrki Lahtonen Mar 19 '13 at 20:29
  • Yes I have but I'm unsure on how to use it. – Michael Mar 19 '13 at 20:33
  • Ok. One way to use it is to first find integers $e_1$ and $e_2$ such that $e_1\equiv1\pmod{14}$ and $e_1\equiv 0\pmod{29}$ (with $e_2$ reverse the roles of the two moduli). Here we see right away that $e_1=29$ works. What about $e_2$? (Hint: look at negative multiples of $14$). – Jyrki Lahtonen Mar 19 '13 at 20:38
  • Then after you have found both $e_1$ and $e_2$, then you can see right away that $$x=ae_1+be_2$$ will be congruent to $a$ modulo $14$ and congruent to $b$ modulo $29$. The CRT says that there is only one residue class modulo $29\cdot 14$ that works, and ou have just found it! – Jyrki Lahtonen Mar 19 '13 at 20:40
  • dude just do your homework. Also the answer to this question is literally posted on learn. –  Mar 20 '13 at 11:05

4 Answers4

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Hint: use the Chinese Remainder Theorem (CRT)

We have pairwise prime moduli, so it can be directly applied.

And see similar posts as to how it can be applied to various systems of linear congruences:

@Jyriki's comments are very helpful, so I won't duplicate in my post the hints and pointers made in the comments. But feel free to check back/comment with your progress, and/or if/where you might be getting stuck.

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amWhy
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  • From reading those questions, I got to gcd(1,14)=6 and gcd(1,29)=24 – Michael Mar 19 '13 at 20:55
  • Well, no need now, I guess, to help you; as you've been given the answer. I just hope you know how Marvis arrived at $(22, 10)$, and understand the CRT well enough to apply it in an exam. If you'd like to know HOW to use it, let me know. – amWhy Mar 19 '13 at 21:02
  • Yes I would like to know how he got to 22,10 which I assume is that he just guessed? – Michael Mar 19 '13 at 21:05
  • That's the problem with just getting an answer: you don't necessarily know why $(22, 10)$ is a candidate, but trust that it is. You've accepted that answer, so perhaps Marvis will elaborate. I try to help users know what they need to know so they can use it; just providing solutions for users seldom helps, in the long run. – amWhy Mar 19 '13 at 21:08
  • Would it happen to have to do something with using EEA? – Michael Mar 19 '13 at 21:10
  • Yes, exactly, Michael! – amWhy Mar 19 '13 at 21:12
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Applying Easy CRT (below), noting that $\rm\displaystyle\, \frac{-18}{29}\equiv \frac{-4}{1}\ \ (mod\ 14),\ $ quickly yields

$$\begin{array}{ll}\rm x\equiv \ \ 6\ \ (mod\ 14)\\ \rm x\equiv 24\ \ (mod\ 29)\end{array}\rm \!\iff\! x\equiv 24\! +\! 29 \left[\frac{-18}{29}\, mod\ 14\right]\!\equiv 24\!+\!29[\!-4]\equiv -92\equiv 314\,\ (mod\ 406) $$

Theorem (Easy CRT) $\rm\ \ $ If $\rm\ m,n\:$ are coprime integers then $\rm\ n^{-1}\ $ exists $\rm\ (mod\ m)\ \ $ and

$\rm\displaystyle\quad \begin{eqnarray}\rm x&\equiv&\rm\ a\ \ (mod\ m) \\ \rm x&\equiv&\rm\ b\ \ (mod\ n)\end{eqnarray} \! \iff x\ \equiv\ b + n\ \bigg[\frac{a\!-\!b}{n}\ mod\ m\:\bigg]\ \ (mod\ mn)$

Proof $\rm\ (\Leftarrow)\ \ \ mod\ n\!:\,\ x\equiv b + n\ [\cdots]\equiv b,\ $ and $\rm\,\ mod\ m\!:\,\ x\equiv b + (a\!-\!b)\ n/n\: \equiv\: a\:.$

$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ mn)\ $ since if $\rm\ x',x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ therefore $\rm\ m,n\ |\ x'-x\ \Rightarrow\ mn\ |\ x'-x\ \ $ since $\rm\ \:m,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = mn\:.\ \ $ QED

Remark $\ $ I chose $\rm\: n,m = 29,14\ $ (vs. $\rm\, 14,29)\:$ since then $\rm\:n \equiv 1\,\ (mod\ m),\:$ making completely trivial the computation of $\rm\,\ n^{-1}\ mod\ m\,\ $ in the bracketed term in the formula.

Math Gems
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For your problem, note that we want $$x = 14k_1 + 6 = 29k_2 + 24$$ Hence, we want to find $(k_1,k_2)$ such that $$29k_2 - 14k_1 = -18 \,\,\,\,\, (\clubsuit)$$ First note that if $(k_1^0,k_2^0)$ are solutions, then $(k_1^0 + 29 n, k_2^0 + 14n)$ are solutions for all $n \in \mathbb{Z}$. Further, these are all the only possible solutions since $\gcd(14,29) = 1$.

The Chinese remainder theorem essentially generalizes the above two lines and you can read more about it from the links @amWhy has posted.

Hence, the goal is to find/guess at-least one solution $(k_1,k_2)$ satisfying $(\clubsuit)$. In your case, $(22,10)$ is one candidate. Hence, all the solutions are of the form $$(k_1,k_2) = (22+29n,10+14n)$$ This means $$x = 14(22+29n) + 6 = 314+406n$$

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I usually solve the two sets of equations $$ \begin{align} x_{14}&\equiv1\pmod{14}\\ x_{14}&\equiv0\pmod{29} \end{align}\tag{1} $$ and $$ \begin{align} x_{29}&\equiv0\pmod{14}\\ x_{29}&\equiv1\pmod{29} \end{align}\tag{2} $$ and get a solution to the given equations with $6x_{14}+24x_{29}$.

The Euclid-Wallis Algorithm yields $$ \begin{array}{r} &&2&14\\ \hline 1&0&1&-14\\ 0&1&-2&29\\ 29&14&1&0\\ \end{array}\tag{3} $$ whose last two columns say $29(1-14k)+14(-2+29k)=1$.

From $(3)$, we get one solution $x_{14}=29(1)$ and $x_{29}=14(-2)$. Thus, $6x_{14}+24x_{29}=-498$.

Solutions are unique $\bmod$ $\mathrm{lcm}(14,29)=406$, so the general solution is $$ 314\pmod{406}\tag{4} $$

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robjohn
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