I have proven that all finite abelian simple $K_3$-groups can be characterised by their orders and the sum of all the numbers of Sylow subgroups.
More precisely, I get the following consequence:
Let $S$ be a finite nonablian simple $K_3$-group with order $p^aq^br^c$, where $p,q,r$ are distinct primes, in other words $S$ is a finite abelian simple group with exactly three distinct prime divisors, i.e., $S\in\{A_5, A_6, PSL(2,7), PSL(2,8), PSL(2,17), PSL(3,3), PSU(3,3), PSU(4,2)\}$. If $|G|=|S|$ and $|Syl(G)|=|Syl(S)|$, then $G\cong S$, where $|Syl(S)|=n_p(S)+n_q(S)+n_r(S)$.
The proof is not hard, as all finite non-abelian simple $K_3$-groups can be charaterised by their orders of Sylow normalizers, via work of other authors.
So I get the following conjecture:
Conjecture: Let $S$ be a finite non-abelian simple group. If $|G|=|S|$ and $|Syl(G)|=|Syl(S)|$, then $G\cong S$
Moreover, I have the following guess:
Let $S$ be one of finite non-abelian simple $K_3$-groups. If $|G|=|S|$ and $G$ is not simple, then $|Syl(G)<|Syl(S)|$.
(I have checked $S\in\{A_5, A_6, PSL(2,7), PSL(2,8), PSL(2,17), PSL(3,3)\}$.)
I do not know whether above guess holds in general. One reason is that there are two non-isomorphic simple groups of order 20160, and $|Syl(A_8)|<|Syl(PSL(4,3))|$. Do we have if $|G|=20160$ and $G$ is not simple, then $|Syl(G)<|Syl(A_8)|<|Syl(PSL(4,3))|$? I have not checked this.
May someone give me some math-software help in calculating a system of equations over non-positive integers like below: $$n_p\equiv 1~(\rm mod~p), n_q\equiv 1~(\rm mod~q), n_r\equiv 1~(\rm mod~r),$$ $$n_p|q^br^c, n_q|p^ar^c, n_r|p^aq^b, n_p+n_q+n_r\geq |Syl(S)|,$$ where $S$ is one of finite abelian simple $K_3$-groups and $|S|=p^aq^br^c$.