Find $x,y,z$ such that:
$$ 6x+15y+20z=17 $$
I found this question in Hua Loo Keng's Number Theory under the Greatest Common Factor and Least Common Multiple section.
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2x,y,z integers? positive? – Paolo Leonetti Jun 27 '17 at 12:43
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1@PaoloLeonetti, they cannot all be positive. – lhf Jun 27 '17 at 12:59
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@lhf I know: they can also potatoes, as far as I can read – Paolo Leonetti Jun 27 '17 at 13:00
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3This one is easy: since $x=y=1$ and $z=-1$ gives a value of $1$, then… – Lubin Jun 27 '17 at 13:02
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Hint $\ $ Scale $, 6 + 15 - 20 = 1,$ by $,17., $ See my answer in the linked dupe for how to do it simply by the extended Euclidean algorithm. – Bill Dubuque Jun 27 '17 at 15:11
2 Answers
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The numbers of the form $6x+15y$ are exactly the multiples of $3=\gcd(6,15)$.
So, first solve $3t+20z=17$ and then solve $3t=6x+15y$ or $t=2x+5y$.
Actually, it is easier in this case to start at the other end.
The numbers of the form $15y+20z$ are exactly the multiples of $5=\gcd(15,20)$. So, first solve $6x+5u=17$ and then solve $5u=15y+20z$ or $u=3y+4z$.
The solution $x=2,u=1$ for $6x+5u=17$ stands out. So does the solution $y=-1, z=1$ for $u=3y+4z$.
lhf
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Including answer of lhf, I saw that $x\equiv 2(mod~5)$ or $x={2,7,12,17,\cdots}$
For each $x$ we get $u$ respectively as $u=1,-5,-11,-17,\cdots $
For each $u, y=-u, z=u$ , satisfy the equation except $u=y=1, z=-1$
So the final answer is $(x,y,z)=(5k+2,6k-1,1-6k)$ for $k=1,2,3,\cdots$ and one left is $(x,y,z)=(2,-1,1)$
Now it is finished.
Atul Mishra
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