Hint $\ $ The common remainder $\rm\,r \le 3,\,$ being a remainder mod $4.\,$ Since both $4$ and $5$ divide $\rm\:x-r\:$ so too does their lcm $20.\:$ Thus $\rm\:x = 20\,k + r,\:$ for $\rm\:0\le r\le 3,\:$ hence $\rm\:x\ne 24.$
Remark $\ $ This is a trivial constant case of CRT = Chinese Remainder. If $\rm\:x\equiv r\pmod{n_k}$ has a constant residue $\rm\,r\,$ for all moduli, then, as above, we infer $\rm\:x \equiv r\pmod n,\,$ for $\rm\:n = lcm\,\{n_k\}.\,$ Further, if $\rm\,r\,$ denotes a remainder (vs. congruence class), $ $ so $\rm\, 0\le r < n_k,\,$ then $\rm\: x = j\,m + r,\:$ where $\rm\:0\le r < n_0,\,$ for $\rm\:n_0 = min\, \{n_k\},\,$ and some $\rm\,j\in\Bbb Z.$
So that's how to generally "predict the values". However, for test problems like this, it is usually quicker to brute-force search for the non-solution, since each test takes at most a few seconds.