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I have been told that it is easier to solve the below 2 questions using Diophantine equations instead of simply trial and error.

1) Find the smallest positive integer which, when divided by 6, gives a remainder of 1 and when divided by 11, gives a remainder of 6.

2) What is the least positive integer which has remainders of 1, 1 and 5 when divided by 3, 5 and 7 respectively?

In terms of my progress, I have attempted to generalize these questions by for example the first question:

Since the integer has a remainder of 1 when divided by 6, it must be of the form 6m + 1

By the way, I have no prior knowledge of Diophantine equations and would appreciate both a brief introduction as well as an explanation of how to apply this knowledge to the above questions.

Thank you :)

mvw
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anonymous
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1 Answers1

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One should maybe say that using this approach replaces the guesswork by some modeling and applying a solution algorithm, so it is more systematic.

Problem 1): From the equations for the divisions $$ u = 6x+1 =11y+6 $$ we get a Diophantine linear equation $$ 6x-11y=5 $$ which is the name for a linear equation with integer coefficients $$ a x + b y = c $$ where one looks for integer solutions $x$, $y$. It has either no solution or infinite many solutions, depending on if $$ \gcd(a,b) \,\rvert\, c $$ where $\gcd(a,b)$ is the greatest common denominator of $a$ and $b$.

Here $\gcd(6,-11)=1$ and $1$ divides $5$, so we have solutions.

A particular solution can be obtained by applying the extended Euclidean algorithm.

$$ \begin{array}{|c|c|c|c|c|} \hline i & q_{i-1} & r_i & s_i & t_i \\ \hline 0 & & 6 & 1 & 0 \\ \hline 1 & & -11 & 0 & 1 \\ \hline 2 & 6:-11 = 0 & 6- 0\cdot (-11) = 6 & 1-0\cdot 0=1 & 0-0\cdot 1=0 \\ \hline 3 & -11:6=-1 & -11 - (-1)\cdot 6 = -5 & 0 -(-1)\cdot 1=1 & 1-(-1)\cdot 0=1\\ \hline 4 & 6:-5=-1 & 6 - (-1) \cdot (-5) = 1 & 1-(-1)\cdot 1=2 & 0-(-1)\cdot 1=1\\ \hline 5 & -5:1=-5 & -5 - 1 \cdot (-5) = 0 & 1-(-5)\cdot 2=11 & 1-(-5)\cdot 1=6\\ \hline \end{array} $$

The row for $i=4$ gives $\gcd(6,-11) = 1$, $s = 2$, $t = 1$. Checking: $6\cdot 2 + (-11) \cdot 1 = 1$, OK.

This solves $$ a s + b t = \gcd(a,b) = 1 \Rightarrow a (5s) + b (5t) = 5 $$

Thus $(x,y)=(5\cdot 2, 5\cdot 1) = (10,5)$ is a particular solution.

The general solution is all solutions of the homogenous equation $$ a x + b y = 0 = a x - t ab + t ab + b y = a(x-tb) + b (y + at) $$ for $t \in \mathbb{Z}$ plus one particular solution.

Here this gives $(10+11t,5+6t)$ for $t \in \mathbb{Z}$ and

$u = 6(10+11t)+1=61-66t$, the smallest positive one is $61$.

Problem 2): The second example gives $$ u = 3x+1 = 5 y + 1 = 7 z + 5 $$ So we could try $$ 3x-5y=0 \wedge 3x+1=7z+5 $$ which gives $$ y=(3/5)x \wedge 3x-7z=4 $$ This leads to $(x,z)=(6+7t,2+3t)$ and $y = (3/5)(6+7t) = (18+21t)/5$ which is integer for $t=2$, giving $(x,y,z)=(20,12,8)$ and $u=61$ again.

WA seems to think so as well (link).

mvw
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