Does Euler totient function gives exactly one value(answer) or LEAST calculated value(answer is NOT below this value)? I was studying RSA when came across Euler totient function. The definition states that- it gives the number of positive values less than $n$ which are relatively prime to $n$.
I thought I had it, until I came across this property:-
Euler Totient function is multiplicative function, that is: $\varphi(mn) = \varphi(m)\varphi(n)$
Now, if $p$ is a prime number, $\varphi(p)=p-1$. 
Putting values of $p$ as 11 and 13 one by one,
$$\varphi(11)=10$$
$$\varphi(13)=12$$
Applying above stated property,
$$\varphi(11\cdot 13)=\varphi(11)\varphi(13)$$
$$\varphi(143)=12 \cdot 10$$
$$\varphi(143)=120$$
Is it correct? Does that mean we have $23$ values between $1$ and $143$ which are not relatively prime to $143$? Sorry if its something basic I'm missing. I'm not some genius at maths and came across this during study of RSA Algo. Thanks.

For some intuition about why $n$ and $m$ must be relatively prime, consider that, for $N=p_1^{a_1}p_2^{a_2}...p_n^{a_n}$,
$$\varphi(N)=N \left (1-\frac{1}{p_1}\right) \left (1-\frac{1}{p_2}\right)...\left (1-\frac{1}{p_n}\right)$$
And for $M=p_m^b$,
$$\varphi(M)=M \left (1-\frac{1}{p_m}\right)$$
If $p_m$ is not one of $p_1,p_2,...,p_n$ then $NM$ has $1$ more unique prime factor than $N$, and
$$\varphi(NM)=NM \left (1-\frac{1}{p_1}\right) \left (1-\frac{1}{p_2}\right)...\left (1-\frac{1}{p_n}\right) \left(1-\frac{1}{p_m}\right)=\varphi(N)\varphi(M)$$
But if $p_m$ is one of $p_1,p_2,...,p_n$, then $NM$ the same number of unique prime factors as $N$, and
$$\varphi(NM)=NM \left (1-\frac{1}{p_1}\right) \left (1-\frac{1}{p_2}\right)...\left (1-\frac{1}{p_n}\right) \ne\varphi(N)\varphi(M)$$
