7. (10 points) The teacher is doing calculation practice with Jiajia, Fangfang, and Mingming. The teacher first gives each of them a number, and then asks them to each pick 3 cards with numbers on them. Jiajia picks 3, 6, 7, Fangfang picks 4, 5, 6, and Mingming picks 4, 5, 8. The teacher then asks them to each multiply two of the numbers on their cards, and then add the number the teacher initially gave them. If the numbers the teacher initially gave them are 234, 235, 236 respectively, and they all calculate correctly, how many different numbers can they possibly get?
【Solution】Solution: According to the analysis, the products Jiajia can obtain are:
$$
3 \times 6=18,3 \times 7=21,6 \times 7=42 ;
$$

The products Fangfang can obtain are: $4 \times 5=20,4 \times 6=24,5 \times 6=30$; the products Mingming can obtain are: $4 \times 5=20,4 \times 8=32,5 \times 8=40$, so the numbers Jiajia can obtain are: $234+18=252,234+21=255,234+42=276$; the numbers Fangfang can obtain are: $235+20=255,235+24=259,235+30=265$; the numbers Mingming can obtain are: $236+20=256,236+32=268,236+40=276$; in summary, a total of 7 different numbers can be obtained.

Therefore, the answer is: 7