Finite numbers that cannot be represented with $4m+7k$ using complete induction I need to prove that there exist finite amount of numbers that cannot be represented with $n=4m+7k~ |~ m,k \in \mathbb{N}$ .  
Starting:  
We say that $A = \{n \in \mathbb{N} | n=4m+7k \geq 18   ~~~ m,k \in \mathbb{N} \} \cup \{1,2, \dots 17\}$ 
We say that:  
$18,10,20,21 \in A$ (by hand)  
And assume each number between $18 \leq m \leq n-1 | n\geq 22$ are in A.  
And so:  
$n-4 \geq 18$ because $n\geq22$ 
We know that $n-4 \in A$
So: $n-4 = 4m +7k \rightarrow  n = 4m+4+7k \rightarrow n=4(m+1)+7k$
And we proved that $n \in A$
And thus $A = \mathbb{N}$   an infinite set and $\{n \in \mathbb{N} | n=4m+7k \geq 18   ~~~ m,k \in \mathbb{N} \}$ is infinite.
so: $\{1...17\}$ is finite set. QED.  
I am not sure about this proof because it seems "specific" about $n-4$, and what about $n-3$?   
I would appreciate your help!
 A: This follows from the Chinese Remainder Theorem, but here is a cool way that I like to teach younger children of realizing this fact. Suppose we have a number $n = 4m_1 + 7k_1$. Note that if $k_1 \geq 1$, then we can immediately conclude that $n + 1$ is expressible as $4m + 7k$, as we can "exchange" one of the sevens for two fours, thereby constructing $n + 1$. That is $n + 1 = 4(m_1 + 2) + 7(k_1 - 1)$. 
With these facts, we now notice that if $n = 4m + 7k$ with $k \geq 3$, then we can construct $n + 1$, $n + 2$, and $n + 3$ just by trading in our sevens one at a time for two fours. Of course, $n + 4$ is easily constructable as it is just $4 (m + 1) + 7k$, and again we can begin the exchanging process to construct the next three natural numbers (since again $k \geq 3$ and hence $n + 4$ has $3$ sevens to "exchange"). This implies that every number after $7 \cdot 3 = 21$ is constructable, so indeed there are only finitely many natural numbers that are not of form $4m + 7k$.
Note that this little heuristic works for any two relatively prime numbers $p$ and $q$. That is, there are finitely many numbers not expressible as $mp + kq$ for such $p$ and $q$. 
