Suppose you have N samples of a sine wave of frequency $f_0$ sampled at $f_s$.
$$ x[n] = A\sin[2\pi(f_0 /f_s)n] $$
I think it's easier to look at exponentials. The reasoning is exactly the same, but if a sine or cosine were used, we would need two exponentials. So first let's say that $$x[n]=e^{j2\pi\frac{f_0}{fs}n} $$ Then the DFT is $$\begin{align} X_N[k]&=\sum_{n=0}^{N-1}e^{-j\frac{2\pi}{N}kn}e^{j2\pi\frac{f_0}{fs}n} \\ &=\sum_{n=0}^{N-1}e^{-j2\pi n\left ( \frac{k}{N}-\frac{f_0}{f_s} \right)} \\ &=\frac{1-e^{-j2\pi N\left(\frac{k}{N} -\frac{f_0}{f_s}\right)}}{1-e^{-j2\pi \left(\frac{k}{N} -\frac{f_0}{f_s}\right)}} \mbox{, where } k=0,\ldots,N-1 \end{align}$$
For this to be a delta, we need the following condition to be satisfied: $$ N\left(\frac{k}{N}-\frac{f_0}{f_s}\right) \in \mathbb{Z} \Rightarrow k-N\frac{f_0}{f_s} \in \mathbb{Z} $$
With this condition, the numerator is always zero. The denominator will be zero if $$ \frac{k}{N}-\left.\frac{f_0}{f_s}\right\vert_{k=k_0}=0 \mbox{, or } k_0=N\frac{f_0}{f_s}$$
At that particular $k_0$, you will need to use L'Hospital's Rule. This will give you the complete DFT as $$X_N[k]=\begin{cases} N &\mbox{ if } k=N\frac{f_0}{f_s} \\ 0 &\mbox { else} \end{cases} \mbox{, } k=0,\ldots,N-1$$
So what conditions did we impose to get this result? Just that $k-N\frac{f_0}{f_s} \in \mathbb{Z}$. Since $k$ is already an integer, we need only check $N\frac{f_0}{f_s}$. Define $f_s=af_0$. Therefore $a$ is the number of points in a cycle. Then we need $$\frac{N}{a} \in \mathbb{Z}$$ This can only be satisfied if $a$ is an integer, and $N$ is an integer multiple of $a$.
Putting all this together, if the sampling frequency is an integer multiple of the input's frequency, and $N$ is an integer number of points in a cycle of the sampled exponential, then the DFT of $x[n]$ will be a delta with height $N$.
In reverse order,
Yes the sampling frequency needs to be a multiple of $f_0$.
The number of cycles of the sinusoid in $N$ samples needs to be an integer, not necessarily a prime number.
If you do 3. and 2., all the energy will be concentrated in two bins (corresponding to $100$ Hz and $-100$ Hz.
Be careful, though. If you have a sinusoid $x(t)=\sin(2\pi (100) t)$ at 100 Hz and you sample it at 1000 Hz, you will have $10$ samples per cycle of the waveform. With $x[0] = 0$ and $x[9] \neq 0$, you can do an "FFT" of length $10$ if you like, and all the energy will be in $X[1]$ and $X[9]$. I put FFT in quotation marks because at such short lengths, a plain vanilla brute-force DFT calculation might be almost as fast. Note also that $x[10] = 0$ is not part of the DFT calculation. Alternatively, you can use $N = 10n$ samples and do an FFT of length $1000$ (say) on samples $x[0]$ through $x[999]$. In this case, the energy will all be in $X[10]$ and $X[1014]$. Regardless of what some other answers will likely tell you,
DO NOT zero-pad the length $1000$ FFT to length $1024$ to get an "efficient" implementation.
If you do follow the advice on zero-padding, your length-$1024$ FFT will NOT have the energy concentrated in two bins as you desire.
For any length of FFT and sample rate Fs, you can calculate the entire set of (positive) frequencies for which a pure sinusoid will produce a delta function in the FFT result. They are:
k * Fs / FFTLength
for integer k = 0 to infinity. (Or to FFTLength/2 if you don't want aliases.)
Any other frequency sinusoid will not produce a delta, but instead a sampled periodic Sinc or Dirichlet function (conjugate mirrored for strictly real input). So you can answer your own 3 questions by process of elimination.
