Frobenius coin problem Suppose that you only have coins worth, say 3 and 5 euros. According to Sylvester result we can find the Frobenius nr $g(3,5)=15-3-5=7$ so 7 is the largest integer that cannot be written as $a_{1}k_{1}+a_{2}k_{2}$ for $k_{1},k_{2}\in\mathbb{N}$ and $a_{1},a_{2}$ are the values of these coins.
a) how do you pay 8€,9€ and 10€ with these coins?
b)use a) to show that it is possible to pay all amounts that are greater than 10€ with the coins 3€ and 5€.
c) show that it is impossible to pay the amount of 7€ with these coins.
I am afraid I do not understand 100% the whole idea behind the Frobenius numbers.
a) can we just take 3€+5€=8€ and 3€+3€+3€=9€ and 5€+5€=10€ this seems suspicious of how easy it is....
b)do I have to use both coins? or just 3€ or 5€?
11€=3€+5€+3€
12€=3€+3€+3€+3€
13€=3€+5€+5€
14€=5€+3€+3€+3€
.
.
c) if we could pay 7€ with these coins we could have written 
$7€=k_{1}5€+k_{2}3€$ but this is impossible as $k_{1},k_{2}\in\mathbb{N}$
can someone please explain to me what should be done in this exercise and how?
 A: For part c, consider the three cases:


*

*No €5 coin is used

*Exactly one €5 coin is used

*At least two €5 coins are used


In each  case, you can easily show that the total can't be €7.
A: Your answer to (a) is correct.  Your answer to (b) needs to have induction, three dots is insufficient.  Your answer to (c) needs to go into a lot of detail about the word "impossible"; why is this so?
A: Your approach to (c) can be made to work. You have the Diophantine equation $3x+5y=7$. One solution is $x=-1,y=2$, so the general solution is 
$$\left\{\begin{align*}
x&=-1+5k\\
y&=2-3k\;.
\end{align*}\right.$$
Since we require that $x\ge 0$, we must have $k\ge 1$, but then $y\le-1<0$, so there is no solution in non-negative integers.
However, the numbers are so small that it’s easier to examine cases, unless you’re very comfortable with solving linear Diophantine equations. Since $3+5>7$, you clearly cannot use both denominations to make $7$. But $7$ is not a multiple of $3$, so you can’t make it using only $3$’s, and it’s not a multiple of $5$, so you can’t make it using only $5$’s. Thus, you can’t make it at all.
For (b) you really do need a proof by induction. For your induction step try to prove that if you can make $n,n+1$, and $n+2$, then you can make $n+1,n+2$, and $n+3$; do you see why that would give you the desired result once you know how to make $8,9$, and $10$?
