Show that any integer $n>7$ can be written as the sum of $3$s and $5$s exclusively 
Show that any integer $n>7$ can be written as the sum of $3$s and $5$s exclusively, 

i.e., 
$$
8= 5+3 \\
9=3+3+3  \\
10 = 5+5 \\
11 = 5+3+3 \\
12 = 3+3+3+3 
$$
So I've started in a couple directions without progress.  I think it makes sense to write
$$n=3x+5y$$
for $x,y \geq 0$, observing that either $3\mid n$ or $5\mid n$.        
It also seems that if you are at $n$, you can increment to $n+1$ by replacing a $5$ with two $3$s.  Or you can take $n-1$ and replace a $3$ with a $5$.  In this way, it seems like you should be able to keep incrementing after $n=8$, but these ideas aren't formalizing into anything...
 A: Here is a way to do this using your ideas:
1) As you have shown, 8 can be written in this way.
2) Let $n$ be in $N$ with $n\ge8$ and $n=3x+5y$ where $x,y\ge0$.
a) If $y\ge1$, $\;n+1=3(x+2)+5(y-1)$.
b) If $y=0$, then $3x=n\ge8\implies x\ge3$, so $\;n+1=3(x-3)+5(y+2)$.
Therefore every integer $n\ge8$ can be written 
in this form by induction.
A: In general, if $a$ and $b$ are two coprime positive integer, every integer $\geq (a-1)(b-1)$ can be written as $ja+kb$ with $j,k\in\mathbb{N}$. This is also known as Frobenius coin (change) problem, see also here.
A: Hint:
You have an expression for $8, 9$, and $10$.
Now if $n$ can be written as such a sum, what can you say about $n + 3$?
A: You're on the right track.
Set $$k = \left\lfloor \frac{n}{5} \right\rfloor.$$ Then,


*

*If $n$ is a multiple of 5, set $x = 0$ and $y = k$ and you're done.

*If $n \equiv 1 \pmod 5$, set $x = 2$ and $y = k - 1$.

*If $n \equiv 2 \pmod 5$, set $x = 3$ and $y = k - 2$.

*If $n \equiv 3 \pmod 5$, set $x = 1$ and $y = k$.

*If $n \equiv 4 \pmod 5$, set $x = 3$ and $y = k - 2$.


Given the requirement that $x \geq 0$, $y \geq 0$, these five settings work provided $k > 2$.
This is not the most efficient way, but it is sufficient and adequate. You might be able to optimize this by changing the denominator to 3 in the fraction to determine $k$, then coming up with the three settings of $x$ and $y$.
A: $2*3 - 5 = 1$
$2n*3 - 5n = n$.
$3(2n - 5k) + 5(3k - n)= n$.
To assure that $2n - 5k \ge 0$ and $3k - n > 0$...
If $n = 3m - r; r = 0, 1,2$ then $k$ can be anything equal or greater than $m$ so long as $2n - 5k \ge 0$ i.e. $6m - 2r - 5k \ge 0\implies k \le 6m/5 - 2r/5= m + \frac{m-2r}5$.  
So long as $m \ge 4$ we will always be able to find such $k$.  i.e. so long as $n \ge 3*4 -2 = 10$.
If $m = 3$ we'll be able find such $k$ if $r \le 1$ i.e. if $n = 8$ or $9$.
We will not be able to find any such $k$ for $n =7$  (where $m = 3; r= 2$).
=======
Or another way:
If we can find $n = 3a + 5b$ we can find $n+1 = 3(a-3) + 5(b+2) = 3(a+2) + 5(b-1)$ so long as either $a \ge 3$ or $b \ge 1$.
If we have $a + b \ge 2$ and $n=3a+5b$ we can find $n+1$ and in doing so either $a$ will increase by $2$ and $b$ (which was at least $1$) decrease by $1$ and $(a+2) + (b-1)\ge 2$  or $a$ (which was at least $3$) will decrease by $3$ but $b$ will in by $2$ so $(a - 3) + (b+2)\ge 2$.
Thus by induction if $n = 3a + 5b; a+b \ge 2$ then $n+1 = 3a' + 5b'; a'+b' \ge 2$.
Base case: $n = 8 = 3*1 + 5*1$.  So is possible for all $n \ge 8$.
A: Since $8=5+3$ we consider those $n$'s for which $n\ge 9$.


*

*If $n=3k$ then we are done. 

*If $n=3k+1$ then observe that $3k+1=3(k-3)+2\cdot 5$.

*If $n=3k+2$ then observe that $n=3k+2=3(k-1)+ 5$. 
