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I am trying to find the smallest positive integer of the form

$30x+6y+10z$, where $(x,y,z)\in\mathbb{Z}$

However, I do not know where to start. Hints or answers are welcome. Thanks!

SeanC
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4 Answers4

7

Hint: The integer has to be even (can you see why?). $6$ yes. Can you get $4$ or $2$? If you can generate $2$ you're done.

Simon S
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  • I can see why it is even. I can set $x,z=0$ and that is a good enough answer? To get $4$ or $2$ would I use modular arithmetic? – SeanC Feb 21 '15 at 23:24
  • The integer has to be even because the expression equals $2(15x + 3y + 5z)$, twice an integer is even.

    As for generating $4$ or $2$, nothing that fancy is needed. Try some values of $x, y, z$

     – Simon S Feb 21 '15 at 23:25
  • In general, could I take the "equation" modulo the gcd and if it equals $0$ that is the answer? – SeanC Feb 21 '15 at 23:31
  • @SeanC: $10+6=16$ and $16\cdot2=32$ and $32-30=2$. – Lucian Feb 21 '15 at 23:51
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Since $\mathbf Z$ is a PID, it is the gcd of 30, 6 and 10 — 2.

Bernard
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Hint $\ $ The set $\rm\,S\,$ of integers of form $\rm\,a_1\,x_1 + \cdots + a_k x_k,\ x_i\in \mathbb Z,\,$ is closed under subtraction so, by this Lemma, every positive $\rm\,n\in S\,$ is divisible by $\rm\,d = $ least positive $\rm\in S.\,$ Now $\rm\,a_i\in S$ $\,\Rightarrow\,$ $\rm d\mid a_i,\,$ i.e. $\rm\,d\,$ is a common divisor of all $\rm\,a_i,\,$ necessarily greatest: $\rm\ c\mid a_i$ $\Rightarrow$ $\rm\,c\mid d = a_!\,x_1\!+\!\cdots\!+\!a_k x_k$ $\Rightarrow$ $\rm\,c\le d.$

Remark $\ $ This is the general form of Bezout's Identity for the gcd.

Community
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Bill Dubuque
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See that $30 - 10 - 3\cdot 6 = 2 (= 2\cdot 10 - 3\cdot 6 = 2\cdot 6 - 10)$.
Also note that in general there always exists an integer linear combination of $\{x,y,z\}$ such that $$\lambda_1 x+ \lambda_2 y + \lambda_3 z = \gcd(x,y,z)$$ Since $\gcd(30,10,6) = 2 \ne 1$ we know that $2$ is in fact the smallest possible integer. $\gcd(x,y,z)$ will always be the least positive integer of that form in general, this is known as Bézout's lemma.

AlexR
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