More than one private key for RSA In an RSA-encryption scenario, Bob's public key pair $(n, e)$ is $(143, 43)$. An attacker Mallory tries brute-force and comes to $d = 7$ as the private key.
The value of $φ(143) = 120$ is not known to Mallory.
However from $43 \cdot d \equiv 1 \bmod 120)$, one can calculate the first positive element $d = 67$ from multiplicative group $d = 120 + 67n$ and $n \in \mathbb{Z}$
$d = 7$ clearly doesn't fit in that multiplicative group, so how come it can successfully decrypt the encryption?
This question can be summarized: the attacker found a $d$ that did not satisfy $e \cdot d \equiv 1 \pmod{ \phi(n) }$, but it works; what's going on.
It turns out that $e \cdot d \equiv 1 \pmod{ \phi(n) }$ is not necessary (it is sufficient).
The necessary and sufficient conditions are:
$$e \cdot d \equiv 1 \pmod{p-1}$$
$$e \cdot d \equiv 1 \pmod{q-1}$$
If both of these hold, then $d$ will always work [1]; conversely if $d$ always works, then both of these hold.
These two conditions can be summarized as a single relation:
$$e \cdot d \equiv 1 \pmod{\text{lcm}(p-1, q-1)}$$
This $\text{lcm}(p-1, q-1)$ modulus is known as the Carmichael function of $n$.
In the specific example you have, $\text{lcm}(p-1, q-1) = 60$, and we have $7 \cdot 43 \equiv 1 \pmod{60}$, and hence $d = 7$ works

[1]: Assuming $p, q$ are distinct primes.