Question 975412
<font face="Times New Roman" size="+2">


Each engineer will touch those and ONLY those switches where the number of the switch is divisible by the engineer's ordinal number.  Every integer has an even number of even divisors, except perfect squares which have an odd number of even divisors.


If *[tex \Large p\ \in\ \mathbb{Z}] is a divisor of *[tex \Large Z_i], then *[tex \Large \exists\ q\ \in\ \mathbb{Z}\ |\ pq\ =\ Z_i].  But that implies that *[tex \Large q\,|\,Z_i] as well.  Hence, divisors come in pairs, except in the case of a perfect square where there is one value of *[tex \Large p] that divides *[tex \Large Z_{sq}] so that there is a *[tex \Large q] such that *[tex \Large pq\ =\ Z_{sq}] AND *[tex \Large q\ =\ p] creating the odd divisor. 


Therefore all of the switches with non-perfect square numbers will be touched an even number of times, and since they started in the off position, they will end up in the off position.


But the perfect square numbers...


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \