Cracking RSA with small plaintext I read a CTF writeup about cracking 4 primes RSA numbers from here:
Given $p, q, r,$ and $p+q+r$ are prime numbers.

The challenge encrypts the flag with a modulus $N=(p∗q∗r)∗(p+q+r)$
and gives the output $n=pqr, k=p+q+r$. To totally break the
cryptosystem, we would want to find the totient of the modulus
$\varphi(N)=(p−1)(q−1)(r−1)(p+q+r−1)$
but we can simplify this when the encrypted message $m$ is small enough.
If we have $m&lt;k$, we can instead find $\varphi(k)=k−1$, and find $e^{-1}\bmod{\varphi(k)}$, and
solve!

But how can they replace $p*q*r*(p+q+r)$ with only $(p+q+r)$ and somebody explains that part?
We have given $c = m^{17} \bmod (N=pqrk)$ with $k = p+q+r$
We can write this as $$m^{17} =  c + \ell (pqr)k$$ for some integer $\ell$.
Now consider this as $$m^{17} =  c + (\ell pqr)k$$
if $m &lt; k$ you can find the $m$ if not information is lost and the answer is not unique.