5. Xiao Hua plays a certain game, in each round he can play as many times as he wants, each time the score is one of the three numbers: $8, a$ (a natural number), 0. The total score of each round is called the total points of that round. Xiao Hua has once achieved the following total points: $103,104,105,106,107,108,109,110$, and it is known that he cannot achieve the total points of “83 points”. What is $a$?
5.【Solution】$83+8 \times 3=107$, so when the total score is 107, the number of rounds scoring 8 points must be less than 3 (otherwise $83=107-3 \times 8$ can be obtained), i.e., the number of rounds scoring 8 points is $0, 1$ or 2, thus among $107, 107-1 \times 8=99, 107-2 \times 8=91$, one of these three numbers must be a multiple of $\mathrm{a}$.

If 107 is a multiple of $a$, then $a=1$ or 107, but when $a=1$, a total score of 83 can be obtained; when $\mathrm{a}=107$, a total score of 103 cannot be obtained, so this situation is impossible. If 99 is a multiple of $\mathrm{a}$, then $\mathrm{a}=1, 3, 9, 11, 33, 99$.

Since $83=9 \times 3+8 \times 7=11+8 \times 9$, $\mathrm{a}$ cannot be $1, 3, 9, 11$ (otherwise 83 can be obtained).

Since $103=99+14=33+70=2 \times 33+37$, when $a=99$ or 33, a total score of 103 cannot be obtained.

Therefore, this situation is also impossible.
If 91 is a multiple of $\mathrm{a}$, then $\mathrm{a}=1, 7, 13, 91$, since $83=7 \times 5+8 \times 6$, so $\mathrm{a} \neq 7.1 \quad 103=91+12$, so $\mathrm{a} \neq 91$.

Therefore, $\mathrm{a}=13$, and it is not difficult to verify that $\mathrm{a}=13$ meets the requirements.