Numbers which cannot be formed We are given two numbers $a,b$ such that $a<b$. Now we have a set $\{a,a+1,a+2,\ldots, b\}$ (all number between a and b including them). Then, we have to find how many numbers cannot be formed from the above set. The only operation allowed on the set elements is addition.  
Note : We can add these number as many times we want. Only addition allowed on these numbers.
Eg : If numbers are 3 and 5, the set is $\{3,4,5\}$ using these we only cannot make $1,2$.
Can anyone help me with that?
I am not getting any idea. I tried the cases when $b-a = 1$. 

Let us consider the numbers that can be formed by $\alpha-1$ additions:
\begin{align}
\alpha =1 \text{ gives }& \mathcal{A}_1 =\{a,a+1,\ldots, a+(b-a)\}\\
\alpha =2 \text{ gives } &  \mathcal{A}_2 =\{2a,2a+1,\ldots, 2a+2(b-a)\}\\
\alpha =3 \text{ gives } &  \mathcal{A}_3 =\{3a,3a+1,\ldots, 3a+3(b-a)\}\\
&\vdots\\
\text{In general, } \alpha \text{ gives } &  \mathcal{A}_\alpha =\{\alpha a,\alpha a+1,\ldots, \alpha a +\alpha(b-a)\}
\end{align}
Note that $ \mathcal{A}_\alpha$ is a set of consecutive numbers, and the size of the set$| \mathcal{A}_\alpha|=\alpha(b-a)$ increases with $\alpha$. 
Suppose for some $\alpha$, $\mathcal{A}_{\alpha+1}$ is the continuation of $\mathcal{A}_{\alpha}$, i.e., there is no gap between the two consecutive group of numbers that can be formed, then all numbers $\geq \alpha a$ can be formed. 
Therefore,  if the last element of $\mathcal{A}_{\alpha}$ is one less than $\mathcal{A}_{\alpha+1}, i.e.,  \alpha a +\alpha(b-a)+1= (\alpha+1) a\implies \alpha=\frac{a-1}{b-a}$, then all numbers $\geq \alpha a$ can be formed using the set. Thus, the gap between the consecutive groups of numbers that can be formed is the set of numbers that can not be formed:

$$\left\{\alpha b+1,\alpha b+2,\ldots,(\alpha+1)a-1,
> \forall \; 0\leq \alpha < \frac{a-1}{b-a}\right\}$$ can not be formed
  using the given set.

For example, 
\begin{align}
a=3,b=5\implies&\frac{a-1}{b-a}=1 \implies \alpha = 0\\
&\left\{\alpha b+1,\alpha b+2,\ldots,(\alpha+1)a-1, \forall \; 0\leq \alpha < \frac{a-1}{b-a}\right\}=\left\{1,2 \right\}\\
a=3,b=4\implies&\frac{a-1}{b-a}=2 \implies \alpha = 0,1\\
&\left\{\alpha b+1,\alpha b+2,\ldots,(\alpha+1)a-1, \forall \; 0\leq \alpha < \frac{a-1}{b-a}\right\}=\left\{1,2,5 \right\}.
\end{align}
