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Given $x=4a+3 \text{ and } x=7b+6,\;\; x,a,b \in \mathbb N,\;\; x,a,b > 0,\;$ find the minimum value for $x$.

How can I solve this system, given three unknown variables but only two equations?

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    Any motivation behind this? What do $a$ and $b$ stand for? Do they bear any relation?2011-05-31
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    You mention in your comment to fmartin that you are looking for a minimum *positive* value for $x$; you should add that to your question. Are there also constraints on a and b? (e.g., do both a and b have to be positive? Or $\geq$ 0?) If there are any additional constraints (e.g. on a and/or b), edit your post to indicate so...that way, perhaps we can help, and perhaps you can obtain the help you want.2011-05-31
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    @fmartin They bear no relation. The $a$ and $b$ variables are unknown quotients; $3$ and $6$ are remainders when dividing $x$ by $4$ and $7$, respectively.2011-05-31
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    @amWhy They are both positive non-zero numbers. (Obviously integers since it's division with remainder.)2011-05-31
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    If $3$ is the remainder when dividing $x$ by $4$, then you have written it wrong. That would be $x = 4a+3$.2011-05-31
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    @GEdgar Ah, yes. Sorry. I was thinking about something else when I wrote that. :/2011-05-31
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    Seems like we need find the smallest positive integer of the form $4a+3$ and $7b + 6$2011-05-31

5 Answers 5

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If $a$ and $b$ are arbitrary real numbers, there's no such minimum since $4a+\frac{3}{4}$ and $7b+\frac{6}{7}$ both span the whole real line. Are you sure there are no additional restrictions on $a$ and $b$? For instance, if $a,b>0$, then the required minimum $$x=4a+\dfrac{3}{4}=7b+\dfrac{6}{7}$$ is $x=\frac{6}{7}$.

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HInt:

$x = -1 \mod 4$ and $x = -1 \mod 7$.

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If $a$ and $b$ are not related in any way, then you have an indeterminate system which allows infinite solutions for $x$.

EDIT: Since $a\in \mathbb{Z}$, if we consider $a$ to be non-negative, then the minimum possible value of $x$ would be with $a=1$, so $x=4+\frac{3}{4}=\frac{19}{4}$.

EDIT 2: New conditions have been added, $x,a,b\in \mathbb{N}$, therefore there's no answer, since the sum of a natural number ($4a$) plus a non-natural number ($\frac{3}{4}$) is always a non-natural number, and so therefore $x\not \in \mathbb{N}$ against our hypotheses.

EDIT 3: Problem has changed again. $4a+3 \equiv 6 \pmod 7$, then $4a \equiv 3 \pmod 7$, then the smallest $a$ that verifies this is $6$, so $x=27$, $b=3$.

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    Indeed, it's an indeterminate system, but is there a way to find the minimum positive value of $x$?2011-05-31
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    It depends. If $x$ belongs to some set with a smallest positive value (or even better, a set with the property that every subset has a smallest value, such as the integers) then it's possible (in such case it would be $1$, of course). If not, e.g. if $x\in \mathbb{R}$ then it wouldn't be possible (the Archimedean property of the reals guarantees there's no infinitely small real numbers).2011-05-31
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    I mentioned some further restrictions on the unknown values in my question (sorry for not being as exact as possible): $x,a,b \in \mathbb N, x,a,b > 0.$2011-05-31
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Chinese Remainder Theorem .....

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If both $x=4a+\frac{3}{4}$ and $x=7b+\frac{6}{7}$, then you have

$$ 4a+\frac{3}{4}=7b+\frac{6}{7}. $$

Then $a$ will be a function of $b$ or vice-versa. As fmartin says, this will admit an infinite number of solutions.