5. (10 points) Teacher Hu originally has 100 red cards and 100 blue cards. Teacher Hu can exchange 2 red cards for 1 blue card and 1 purple card; or exchange 3 blue cards for 1 red card and 1 purple card. After several exchanges, what is the maximum number of purple cards that Teacher Hu can hold? $\qquad$
【Analysis】According to the problem, we can use the fact that the total score remains unchanged before and after the exchange. Since it is impossible to end up with all purple cards, and 5 points cannot be represented as the sum of several 3s and 4s, $10=3+3+4$, the maximum number of purple cards that can be obtained is: $(700-10) \div 5=138$ (cards).

【Solution】According to the analysis, assuming each blue card is worth 3 points, each red card is worth 4 points, and each purple card is worth 5 points, the total score remains unchanged after each exchange. The total score is: $3 \times 100 + 4 \times 100 = 700$ (points). Since it is impossible to end up with all purple cards,

and 5 points cannot be represented as the sum of several 3s and 4s, $10=3+3+4$, the maximum number of purple cards that can be obtained is: $(700-10) \div 5 = 138$ (cards). The operations can be as follows:
$$
\begin{array}{l}
(100,100,0) \rightarrow(0,150,50) \rightarrow(50,0,100) \rightarrow(0,25,125) \\
\rightarrow(8,1,133) \rightarrow(0,5,137) \rightarrow(1,2,138)
\end{array}
$$

Therefore, the answer is: 138.