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I'm doing discrete math. Been stuck on this problem forever. I need to find a pair of integers $x$ and $y$ such that $$17369x + 5472y = 4.$$ I understand that I need to use the division algorithm. But what after that? If someone can offer a step by step solution, it would be much appreciated!

Thanks guys :)

Juniven Acapulco
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Vimzy
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3 Answers3

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Division algorithm: 5472 divides into 17369 three times with remainder 953: 17369= 3(5472)+ 953 so 17369- 3(5472)= 953

953 divides into 5472 5 times with remainder 707: 5472= 5(953)+ 707 so 5472- 5(953)= 707

707 divides into 953 once with remainder 246: 953= 1(707)+ 246 so 953- 1(707)= 246

246 divides into 707 twice with remainder 215: 707= 2(246)+ 215 so 707- 2(246)= 215

215 divides into 246 once with remainder 31: 246= 1(215)+ 31 so 246- 1(215)= 31

31 divides into 215 six times with remainder 29: 215= 6(31)+ 29 so 215- 6(31)= 29

29 divides into 31 once with remainder 2: 31= 1(29)+ 2 so 31- 1(29)= 2

2 divides into 29 14 times with remainder 1: 29= 2(14)+ 1 so 29- 2(14)= 1

Now we work our way back up that list- Replace that "2" by 31- 29: 29- 14(31- 29)= 15(29)- 14(31)= 1

Replace that "29" by 215- 6(31): 15(215- 31)- 14(31)= 15(215)- 29(31)= 1

Replace that "31" by 246- 1(215): 15(215)- 29(246- 215)= 44(215)- 29(246)= 1

Replace that "215" by 707- 2(246): 44(707- 2(246))- 29(246)= 44(707)- 117(246)= 1

Replace that "246" by 953- 1(707): 44(707)- 117(953- 1(707))= 161(707)- 117(953)= 1

Replace that "707" by 5472- 5(953): 161(5472- 5(953))- 117(953)= 161(5472)- 922(953)= 1

Replace that "953" by 17369- 3(5472): 161(5472)- 922(17369- 3(5472)= 2927(5472)- 922(17369)= 1

That's equal to "1" and we want "4" so MULTIPLY BY 4! 11708(5472)- 3688(17369)= 4.

The original problem was to find x and y such that 17369x+ 5472y= 4. Looking at the previous equation, we see we can set x= -3688 and y= 11708.

user247327
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First step: $17369 = 5472 \times 3 + 953$. So if you write $y = y_1 - 3 x$, the equation becomes $953 x+5472 y_1 = 4$. Continue with $5472 = 953 \times \ldots + \ldots$ and $x = x_1 - \ldots y_1$.

Robert Israel
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  • I did do that, and eventually ended up with 29 = 2(14) + 1 but the remainder is never equal to 4 when I do this. – Vimzy Feb 26 '14 at 02:37
  • After $2 x_3 + 29 y_4 = 4$ you should get $2 x_4 + y_4 = 4$ with $x_3 = x_4 - 14 y_4$. It's easy to solve this: $y_4 = 4 - 2 x_4$ for any $x_4$. Then $x_3 = x_4 - 14 y_4 = 29 x_4 - 56$ etc. – Robert Israel Feb 26 '14 at 02:44
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Using the Extended Euclidean Algorithm

$$\begin{array}{rrr} 17369 & 1 & 0\\ 5472 & 0 & 1\\ 953 & 1 & -3\\ -246 & -6 & 19\\ -31 & -23& 73\\ 2 & \color{#c00}{178} & \color{#0a0}{-565}\\ \end{array}$$

where each above line $\,\ a\ \ b\ \ c\ \,$ means that $\ a = 17369\, b + 5472\, c.\ $ Therefore

$$ 2 \,=\, 17369\cdot \color{#c00}{178}+ 5472(\color{#0a0}{-565})\quad $$

Multiplying the above by $\,2\,$ yields the sought integers $\,x,y.$

The linked post described the algorithm in great detail, in a way that is easy to remember.

Here is another example computing $\rm\ gcd(141,19),\,$ with the equations written explicitly

$$\rm\begin{eqnarray}(1)\quad \color{#C00}{141}\!\ &=&\,\ \ \ 1&\cdot& 141\, +\ 0&\cdot& 19 \\ (2)\quad\ \color{#C00}{19}\ &=&\,\ \ \ 0&\cdot& 141\, +\ 1&\cdot& 19 \\ \color{#940}{(1)-7\,(2)}\, \rightarrow\, (3)\quad\ \ \ \color{#C00}{ 8}\ &=&\,\ \ \ 1&\cdot& 141\, -\ 7&\cdot& 19 \\ \color{#940}{(2)-2\,(3)}\,\rightarrow\,(4)\quad\ \ \ \color{#C00}{3}\ &=&\, {-}2&\cdot& 141\, + 15&\cdot& 19 \\ \color{#940}{(3)-3\,(4)}\,\rightarrow\,(5)\quad \color{#C00}{{-}1}\ &=&\,\ \ \ 7&\cdot& 141\, -\color{#0A0}{ 52}&\cdot& \color{#0A0}{19} \end{eqnarray}\qquad$$

Community
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Bill Dubuque
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  • My textbook gave the answer as 17369(-10588) + (5472)(33608) = 4 is there a reason for this? – Vimzy Feb 26 '14 at 03:04
  • @Vinzy Yes: your textbook did it the hard way, and I did it the easy way (using smallest remainders). If $,(x,y),$ is a solution then so to is $,(x,y) + n (-5472,17369).,$ Choose $,n=2,$ to go from my solution $,(356,-1130),$ to $,(-10588,33608),$ in your textbook. – Bill Dubuque Feb 26 '14 at 03:37