Problem with RSA deciphering I don't quite get the algorithm yet. Sometimes it works and other times it doesn't,so clearly I am overseeing or misunderstanding something. 

I will just write what I did. 
My $N=143$ and has factors $p=11$, and $q=13$.
To determine my second public number: 
$R=(p-1)(q-1)= 10 \cdot12=120$. So the second number can not be a factor of $120$. I figured $e=7$ would be fine. 

I simply want to message $'7'$. So $\mod\frac{7^7}{143}= 6$ will be my message. 

My friend wants to decode it and needs to exponentiate this number by $d$.
$d=\frac{R+1}{e}=121/7$ But this should be natural number right? 

I noticed it doesn't work for $e=9$ as well. Even though it is not a factor of $120$. It does work for $e=11$.

Should $e$ be chosen so that it is a factor of $R+1$?
For textbook RSA, we have

Key-Gen


The modulus $n$ must be a product of two distinct random large primes due to the security, $n = p \cdot q$ 

in your case $n=143 = 11\cdot 13$

For finding the primes, the probabilistic Miller–Rabin primality test, it should be enough. Note that the Miller–Rabin primality test is probabilistic; composite output is always true, prime output has probability defined by the number iterations, $k$; $$\Pr( p\text{ is not prime} ) \ll \frac{1}{4^k}$$ and can be stated as $$\log_2(\Pr( p\text{ is not prime} ))\ll-2k$$ This is a rough calculation, and as noted by fgrieu, the probability is approaching 0 as the size of the number to be tested increases. The FIPS 186-4 table C.3 provides specific numbers for $k$;


for 512 bits gives $k=7$ rounds with $\log_2(\Pr( p\text{ is not prime} ))&lt;-100$, 
for 1024 bits gives $k=4$ rounds with $\log_2(\Pr( p\text{ is not prime} ))&lt;-100$, and
for 1536 bits gives $k=3$ rounds with $\log_2(\Pr( p\text{ is not prime} ))&lt;-100$.

The factors of modulus are $p=11$ and $q =13$
$\varphi(n) = (p-1)(q-1)$, in your case  $\varphi(143) = 10\cdot 12 = 120$,

Actually, we prefer $\lambda(n) = \operatorname{lcm}(p,q)$ and this will give us the smallest private exponent. That can be helpful for signature calculation speed, and actually, one should use the CRT method ( see the last bullet of Key-Gen)

The relation is;
$$\varphi(n)=\lambda(n)\cdot\gcd(p-1,q-1)$$ and this implies that $\lambda(n)| \varphi(n)$
The public exponent $e$ is chosen relatively prime to $\varphi(n)$, so $e=7$ is fine. Normally the $e$ is chosen advance in $\{3, 5, 17, 257, 65537\}$. If the $\gcd(e,\varphi(n)) \neq 1$ then a new modulus is generated.

$(n,e)$ makes the public key to distribute.
The private exponent $d$ is the inverse of $e$ modulo $\varphi(n)$, i.e. $d\cdot  e \equiv 1 \bmod \varphi(n)$, in your case $d=103$. This can be used with the Ext-GCD which result in a Bézout's identity $ e \cdot x + n \cdot k =1$. Take modulus $n$ then $x$ is the inverse of $e$.

$(n,e,d,p,q, d_p, d_q, q_{inv})$ is your private key. One can use CRT to speed up the decryption up to 4 times.


Encrypt


$c = m^e \bmod n$

The $m \in [0,n)$, otherwise after the decryption one will get an equivalence class representative of $m$ less then $n$.


Decrypt


$m = c^d \bmod n = (m^{e})^{d} \bmod n = m^{ed} \bmod n = m$


Example


$m = 7$ then  $c = 7^7 \pmod{143} = 6$
$m = 6^{103} \pmod {143}= 7$




Notes:


There is also multi-prime RSA where the large prime factors of $n$ are mode than 2.  
Textbook RSA is not secure one should never use it without a proper padding scheme. One is the PKCS#v1.5 padding scheme and the other is RSA-OAEP. RSA OAEP has a security proof and PKCS#v1.5 has not. PKCS#v1.5 has many attacks over the years and should not be used.
RSA ( actually any public-key encryption) is not preferable due to the speed. We prefer the hybrid encryption schemes like RSA-KEM for Key Encapsulation Mechanism then encrypt the data with AES-GCM or ChaCha20-Poly-1305 to achieve Data Encapsulation Mechanism, use 256 bit key with AES, preferably.

With this composition of a KEM and a DEM, one can achieve IND-CCA2/NM-CCA2—ciphertext indistinguishability and nonmalleability under adaptive chosen-ciphertext attack.