Ta có
2p+n=52\(\rightarrow\) n=52-2p
mà 1\(\le\)\(\frac{n}{p}\)\(\le\)1,5
\(\rightarrow\) 1\(\le\)\(\frac{\text{52-2p}}{p}\) \(\le\) 1,5
\(\rightarrow\)14,85\(\le\) p \(\le\)17,333
Mà p> 16 \(\rightarrow\) p=17
\(\rightarrow\)n=52-2.17=18
Vậy p=e=17; n=18