How to show that $2x + 3y + 5z$ generate all integers greater than 2

Question

The title says it all. I am having trouble figuring out how to prove that

$$ 2x + 3y + 5z \quad x,y,z \in \mathbb{N}_{\geq 0}$$

generates all the integers greater than $2$. Please only provide hints as I would like to solve it myself. Thank you.

Answer 1

In fact, $2x+3y$ generates all integers greater than $2$.

If $n$ is even, we put $n = 2k + 3(0)$.

If $n$ is odd, we put $n = 2k+1 = 2(k-1) + 3 \cdot 1$.

Answer 2

Hint:

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If $n$ is even, it is of the form $2k$

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All odd numbers are of the form $3+2k$

Answer 3

In this answer, it is shown that

Theorem 1: Suppose $(a,b)=1$ and $c\ge(a-1)(b-1)$. Then $ax+by=c$ has a non-negative solution, that is, one in which both $x$ and $y$ are non-negative integers.

Thus, every integer greater than or equal to $(2-1)(3-1)=2$ can be written as $2x+3y$ for $x,y\ge0$.

Answer 4

You specifically asked for a hint, and nothing more.

Bézout's Identity says that if $m$ and $n$ are two integers with $\gcd(m,n)=d$ then there exist integers $p$ and $q$ for which $pm+qn=d$. For example, if $m=7$ and $n=11$ then $\gcd(7,11)=1$ and $$(-3)(7) + (2)(11) = 1$$