I started to solve this problem by writing out what
would happen to the first ten switches:
S
W
IT
C
H
ES
MATHEMATICIANS
1
2
3
4
5
6
7
8
9
1 ON /
/
/
/
/
/
/
/
2 ON OFF /
/
/
/
/
/
/
3 ON / OFF /
/
/
/
/
/
4 ON OFF / ON /
/
/
/
/
5 ON /
/
/ OFF /
/
/
/
6 ON OFF ON /
/ OFF /
/
/
7 ON /
/
/
/
/ OFF /
/
8 ON OFF / ON /
/
/ OFF /
9 ON / OFF /
/
/
/
/
ON
10 ON OFF /
/ ON
/
/
/
/
I observed that the bulbs that remain ON are the
square numbers – 1, 4 and 9.
I also noticed that to remain ON, a switch had to be
flicked an odd number of times.
For a switch to be flicked an odd number of times, it
had to have an odd number of factors. The only
numbers with an odd number of factors are square
numbers, because one of their factors is repeated
twice.
For example, the factors of 24 are:
1, 2, 3, 4, 6, 8, 12 and 24.
The factors of 25 are:
1, 5, 5 and 25
5 is repeated twice, so 25 has an odd number of
factors, meaning it will remain ON after all the
mathematicians have passed through the room.
10
/
/
/
/
/
/
/
/
/
OFF