Escape the castle:
In room one the key is 13. we got this because we knew that the opposite sides of a dice adds up to seven. therefore we found all the opposible combinations which added up to eight and converted them to their 'oppisite'. we then descovered that all the combinations added up to thirteen.
In room two the key is 9. We found this out because there are ten cards, with the numbers 0-9 written on them, lying down on the table. five cards are facing up, five are facing down.The numbers on the first two cards add to 13, the numbers on the second and third cards add to 9, the numbers on the third and fourth cards add to 11, and the numbers on the fourth and fifth cards add to 16.
If you try all the possible solutions there is only one solution that works;
8+5 - 5+4 - 4+7 - 7+9. since the last number is nine than the key number is nine.
In room three the key is fifteen. we got this because we knew that the diameter was five so therefore the raidius is 2.5. The overlap is one cm due to the fact that the last key number was nine if you minus one from 2.5 you get 1.5. so either side of the over lap is 1.5. when you add the 1.5s on either side it equals three then if you add the 1 overlap distance it makes four. four squared is sixteen minus one is fifteen.
In room four the key is fourteen. we got this by a simple trial and error and after a while we got the corner numbers 6 8 3 7 which add to 24. 24-10=14.
In room five the key is 19. the 2 trays can hold up to 14 tarts each
If the tarts are counted in fours there are three left over.
If they are counted in threes there is one left over.
ta menas that on 1 tray there will be 12 tarts and on tray 2 there will be 4 tarts and 3 left over equaling 19.
the finial problem spells simon.