Many $(\kappa+\alpha)$-strong cardinals below a $(\kappa+\beta)$-strong one for $\alpha < \beta < \kappa$ Let $\lambda$ be an ordinal. A cardinal $\kappa$ is $\lambda$-strong iff there is some inner model $M$ and an elementary embedding
$$
j \colon V \rightarrow M
$$
s.t. $crit(j) = \kappa$ and $V_\lambda \subseteq M$.
I am given the following exercise:
Let $\alpha < \beta < \kappa$ be ordinals, where $\kappa$ is $(\kappa+\beta)$-strong. Then
$$
\{ \mu < \kappa \mid \mu \text{ is } (\kappa+\alpha) \text{-strong} \}
$$
has size $\kappa$.

I'd like to show that
$$
M \models \kappa \text{ is } (j(\kappa)+\alpha) \text{-strong},
$$
but I'm not sure where to begin...
(It seems that I have to "stretch" a $(\kappa, \kappa+\alpha)$-extender $E \in V$ to a $(\kappa, j(\kappa)+\alpha)$-extender $\tilde E \in M$, but how? Is this even the right approach?)

edit:
Consider the case $\alpha =0 $ and recall that a cardinal $\mu$ is $\mu$-strong iff it is measurable.
Now, if $\kappa$ is $\kappa+2$-strong, we may fix a witnessing measure $U$ on $\kappa$. Then $U \in V_{\kappa+2} \subseteq M$ and some basic observations yield
$$
M \models \kappa \text{ is measurable}
$$
Thus for any fixed $\xi < \kappa$ we get
$$
M \models \exists \mu \colon j(\xi) < \mu < j(\kappa) \wedge \mu \text{ is measurable}
$$
(take $\mu = \kappa$) and elementarity yields that there is some measurable $\mu \in (\xi, \kappa)$. This proves that there are $\kappa$ many $\mu$-strong cardinals $\mu < \kappa$ and also suggest that we should be more careful and require something like $\alpha+n < \beta < \kappa$ for some $0 < n < \omega$. (Given the kind of proof I have for the edited question below, the exact value of $n$ is then easily calculated.)
I think that this exercise contains a typo and should read
$$
\{ \mu < \kappa \mid \mu \text{ is } (\mu+\alpha) \text{-strong} \}
$$
instead. I will ask my professor and in the case that this is what he was asking for, I will tidy up my post and provide an answer for future reference.
 A: We can show that if $\kappa$ is $(\kappa+\alpha+1)$-strong, then the set of $(\kappa+\alpha)$-strong cardinals below $\kappa$ has size $\kappa$. First off, as is easy to verify, $M\vDash \kappa\text{ is }(\kappa+\alpha)\text{-strong}$. Let $k: M\prec N$ witness this. Let $e=k\circ j$. Then $e: V\prec N$ witnesses that $\kappa$ is $(\kappa+\alpha)$-strong with target $\lambda\gt j(\kappa)+\alpha$, and let $E$ be the $(\kappa,e(\kappa))$-extender defined by $X\in E_a$ if and only if $X\in Y$ and $a\in e(X)$, where $Y=\{X\in M|X\subseteq [\kappa]^{|a|}\}$. Then $j_E^M: M\prec \text{Ult}_E(M)$ witnesses, in $M$, the $(j(\kappa)+\alpha)$-strongness of $\kappa$, because $e(\kappa)\gt |V_{(j(\kappa)+\alpha)}|^M$ if and only if $k(\kappa)\gt |V_{\kappa+\alpha}|$. To see this, note that $k(\kappa)\gt |V_{\kappa+\alpha}|^M$, because $k(\kappa)\gt |V_{\kappa+\alpha}|^N$, and since $M\vDash k(\kappa)\times V_{\kappa+\alpha}\subseteq N$, $|V_{\kappa+\alpha}|^M=|V_{\kappa+\alpha}|^N$, and since $k(\kappa)\times V_{\kappa+\alpha}\subseteq M$, $|V_{\kappa+\alpha}|^M=|V_{\kappa+\alpha}|$. To establish $E$ is an extender, condition 1. is easy. Coherence follows from the fact that all the functions, if they exist in $M$, are $\Delta_0$. Normality is also simple. Finally, it is easy to establish that the direct limit is well-founded. The rest follows by standard reflection arguments.
All of this aside, I do think it was probably a typo. This process is rather involved (For an exercise), and the result that $\{\mu\lt\kappa|\mu\text{ is }(\mu+\alpha)\text{-strong}\}$ is much simpler and easier. Plus, I didn't even realize you wrote $\kappa+\alpha$ instead of $\mu+\alpha$, which indicates the author might have as well.
