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Is there any mathematical way to find the integer numbers that solve the following equation:

$$217 = (20x+3)r+x$$

hola
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Kumar
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3 Answers3

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HINT:

$$\implies r=\frac{217-x}{20x+3}$$

If integer $d$ divides both the numerator & the denominator; it will divide
$\displaystyle(217-x)\cdot20+(20x+3)\cdot1=4343=43\cdot101$

So, $\displaystyle20x+3$ must divide $4343$ to keep $r$ an integer

Fortunately, the number of divisors are limited namely, $\displaystyle\pm1,\pm4343,\pm43,\pm101$

Again $\displaystyle\pm1,\pm101\equiv\pm1\pmod{20};$ and $\displaystyle-43,-4343\equiv-3\pmod{20}$ unlike $\displaystyle20x+3\equiv3$ which leaves only two options, right?

lab bhattacharjee
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  • thanks @lab, I need x and r integer values, I tried x = 2 and r = 5. and it is solve the equation. but how I can produce these integer numbers using math way? – Kumar Aug 15 '14 at 16:40
  • @Kumar, Not sure about "math way". But $20x+3$ can be $43$ or $4343;$ $$20x+3=43\iff x=?$$ and $$20x+3=4343\iff x=?$$ Check if each of them keeps $r$ an integer – lab bhattacharjee Aug 15 '14 at 16:44
  • (217−x)20+20x+3=4343 , what this mean?? how you get the 4343? – Kumar Aug 15 '14 at 16:53
  • @Kumar, Please don't accept an answer unless fully cleared. If integer $d$ divides integers $a,b;d$ will divide $ax+by$ for integer $x,y$ – lab bhattacharjee Aug 15 '14 at 16:55
  • sorry, but i don't understand how you found 4343 :$ – Kumar Aug 15 '14 at 17:03
  • @Kumar, $\displaystyle(217-x)\cdot20+(20x+3)\cdot1=?$ – lab bhattacharjee Aug 15 '14 at 17:05
  • @Kumar If $d$ divides $217-x$ and $20x+3$ it must divide their GCD. No do polynomial long division: divide $20x+3$ to $217-x$, the reminder is $4343$... That is where $4343$ comes from..... The relevance is the Euclidian algorithm. – N. S. Aug 15 '14 at 17:26
  • I got it, thanks a lot @labbhattacharjee – Kumar Aug 15 '14 at 17:34
  • @Kumar: if an integer $d$ divides $a$ and $b$ then it will also divide any $ax+by$. If $r=\frac{217-x}{20x+3}$ is an integer, $20x+3$ divides $217-x$ and itself, so it also divides $(217-x)\cdot20+(20x+3)\cdot1$, which simplifies to 4343. – Rory Daulton Aug 15 '14 at 18:11
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$$217 = (20x+3)r+x\\ 217=20xr+3r+x$$

Now $$(5x+\frac{3}{4})(4r+\frac{1}{5})=20xr+3r+x+\frac{3}{20}$$

Therefore $$217+\frac{3}{20}=(5x+\frac{3}{4})(4r+\frac{1}{5})$$

Multiplying both sides by $20$ we get $$4343=(20x+3)(20r+1)$$

There are only few ways of writing $4343$ as a product of two integers, in each case solve. Note that the product has to be of terms of the form $1 \pmod{20}$ and $3 \pmod{20}$.

P.S. Here is a faster way to get to the product: $$217 = (20x+3)r+x$$ Note that $x=\frac{(20x+3)-3}{20}$. Therefore $$217 = (20x+3)r+\frac{(20x+3)-3}{20}=(20x+3)(r+\frac{1}{20})-\frac{3}{20}$$

Move $\frac{3}{20}$ on the other size and multiply by $20$.

N. S.
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  • the number should be x = 2 and r = 5, how can have these numbers? – Kumar Aug 15 '14 at 16:43
  • @Kumar The prime factorization of $4343$ is $4343=43*101$. Thus, $4343$ can be written as a product of two integers only in $8$ ways: $1 \cdot 4343, 43 \cdot 101, 101 \cdot 43, 4343 \cdot 1$, respectively the negative ones. Only two of them are of the form $(20x+3) \cdot (20r+1)$, and one of them leads to $r=0$, which I think you seem to want to exclude, which probably means you only want positive solutions ... – N. S. Aug 15 '14 at 17:13
  • @Kumar See the linked post in my answer for a ay to view the factorization as a generalization of completing the square. – Bill Dubuque Aug 15 '14 at 18:58
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Hint $\ $ Completing a square generalizes to completing a product

$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff &&\!\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

Your $\,ad\!+\!bc\,$ is a product of two primes, so LHS factors are highly constrained.

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Bill Dubuque
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