Representation of all numbers that bigger than $12$ by $3x+7y$ proof I found an exercise in my book that requested from me to proof that all numbers that bigger than $12$ can be represented by:
$3x+7y$
They requested an induction proof,and i decided to share my answer with you,just to be sure about it.
Answer:
Base Case:
we will check three base cases,$n=12,13,14$:
1.for $n=12$:
$12=3x+7y \Rightarrow x=4,y=0$
2.for $n=13$:
$13=3x+7y \Rightarrow x=2 , y=1$
3.for n=14:
$14=3x+7y \Rightarrow x=0,y=2$
we can assume that:
$n-1=3x+7y$
$∀n-1>11.$
Induction step:
We know that modulus 3 divide all the numbers to three groups of numbers:


*

*$A=\{n|n≡0(\mod3)\}$

*$B=\{n|n≡1(\mod3)\}$

*$C=\{n|n≡2(\mod3)\}$


From the base case we can assume:
$∀a ∈ A:∃ n=3x$
$∀b ∈ B:∃ n=3x+7y$
$∀c ∈ C:∃ n=3x+14$
From that we can assume that:


*

*if $n ∈ A$ and $n-1 > 11$ then $n-1 ∈ C$.

*if $n ∈ B$ and $n-1 > 11$ then $n-1 ∈ A$.

*if $n ∈ C$ and $n-1 > 11$ then $n-1 ∈ B$.
Is there any mistakes?
Is that a legit induction proof?
 A: This is not necessarily a proof but shows a relationship to finding the form;
$\forall{n}$ s.t. $3|n, n\ge{12}$, then $n=3x,$ for some $x\in\mathbb{N}$
$$n=3x+7\cdot0$$
$$n+1=3(x-2)+7\cdot1$$
$$n+2=3(x-4)+7\cdot2$$
Since 
$$n+1=3x+1=3x-6+7=3(x-2)+7\cdot1$$
and 
$$n+2=3x+2=3x-12+14=3(x-4)+7\cdot2$$
and of course
$$n+3=3x+3=3(x+1)+7\cdot0$$
A: An obvious approach would be to explicitly solve this for $3x + 7y = 12 \text{ to } 18$ th
$$
\begin{array}{c|ll}
3x + 7y  & x & y  \\
\hline
12 & 4 & 0 \\
13 & 2 & 1 \\
14 & 0 & 2 \\
15 & 5 & 0 \\
16 & 3 & 1 \\
17 & 1 & 2 \\
18 & 6 & 0
\end{array}
$$
Now for any number $n \ge 19$ what can you say?
A: We check the correctness for $n=12$ and $n=13$:
$$12=3\cdot4+7\cdot0$$
$$13=3\cdot2+7\cdot1$$
Now let's assume correctness for $n$:
$$n=3x_1+7y_1$$
Now let's prove for $n+1$:
$$n+1=3x_2+7y_2$$
Let's use our assumption:
$$3x_1+7y_1+1=3x_2+7y_2$$
$$1=3x_2-3x_1+7y_2-7y_1/+12$$
$$13=3x_2-3x_1+7y_2-7y_1+3\cdot4$$
$$13=3(x_2-x_1+4)+7(y_2-y_1)$$
Now that's correct, as we checked:
$$13=3x+7y$$
So the theorem is correct under the Induction Axiom.
