Induction problem basics 
For which values of $n \in \mathbb{N}$ can a group of $n ≥ 1$ people be divided into teams, where each team consists of exactly $4$ or $7$ people? Use induction to prove your answer correct.


I just made a long list of $n$ values and saw that after $n=18$, all values of $n$ can be split up in terms of $4$ or $7$. That's about it...how do I start?
Thanks!
 A: Base $n=18,19,20,21$: $$18 = 1\cdot 4+2\cdot 7$$
$$19 = 3\cdot 4+1\cdot 7$$
$$20 = 5\cdot 4+0\cdot 7$$
$$21 = 0\cdot 4+3\cdot 7$$
Now $n\to n+4$:
By I.H. $n= 4a+7b$ for some $a,b\geq 0$ Then $n+4 = 4(a+1)+7b$ and we are done. 
A: We want to show that $$4a+7b=n$$ has a non-negative integer solution for every $n\ge 18$ :
For $18$, we have a solution because of $1\cdot 4+2\cdot 7=18$
Suppose , we have a non-negative integer solution of $4a+7b=n$ , where $n\ge 18$
We have to show that we can solve $4a'+7b'=n+1$ as well.
If $b>0$, then we can take $a'=a+2$ and $b'=b-1$ Then, we have $$4a'+7b'=4a+8+7b-7=4a+7b+1=n+1$$
If $b=0$, then we have $a\ge 5$ because otherwise $4a+7b<18$
Taking $a'=a-5$ and $b'=b+3$ gives $$4a'+7b'=4a-20+7b+21=4a+7b+1=n+1$$
This completes the proof.
A: hint:
we need to check basis scenario (18 = 14 + 4) assume that we proved it for n so there are a,b >= 0:
n = 4a + 7b
now lets look on n+1, if b >= 1 so n+1 = 4(a+2) + 7(b-1) 
otherwise b = 0 and because n >= 18 then a >=5 means n = 4a >= 20 so n+1 >= 21
so n+1 = 4(a-5)+21
