The Great Aviary Chandelier
by agbdmrbirdyface, Aug 13, 2016, 3:08 AM
Oops I realized this blog actually didn't have a post for today!
So, I have a small and somewhat easy problem of my own creation (and it's pretty trash, but we need a problem!)
Solution
So, I have a small and somewhat easy problem of my own creation (and it's pretty trash, but we need a problem!)
agbdmrbirdyface wrote:
agbdmrbirdyface has a very large circular chandelier in his aviary, with 2015 spots for light bulbs. Each of the spots is numbered consecutively from 1 to 2015. He is installing new light bulbs in his chandelier, so at the start, his chandelier has no light bulbs in any of the spots. agbdmrbirdyface wants to install light bulbs in his chandelier according to the following rule: for the nth iteration of this rule, agbd will install n light bulbs in the first n empty lightbulb spots after the n-1th iteration, starting with the first empty lightbulb spot after the n-1th iteration of the rule is complete. After this is complete, he will skip one lightbulb spot. .
agbd will keep applying this rule until all the lightbulb holders are full, even if he has to go back around the circle again to install more light bulbs. If agbd installs the first lightbulb according to the rule in spot number 1, which spot is the last to be filled with a lightbulb?
agbd will keep applying this rule until all the lightbulb holders are full, even if he has to go back around the circle again to install more light bulbs. If agbd installs the first lightbulb according to the rule in spot number 1, which spot is the last to be filled with a lightbulb?
Solution
The number of lightbulb spots that have been either skipped or filled by the rule after its nth iteration can be expressed as
.
We may be able to use this expression to see how many iterations it will take to either skip or fill 2015 spots... and it turns out, by convenience, that the quadratic
has an integer root
.
So, after
iterations, spot number
will be the last skipped lightbulb spot. Then, on iteration
, all the previous skipped spots will then be filled, since
light bulbs will be placed in
empty spots. Hence, the last lightbulb spot to be filled will be spot number
.

We may be able to use this expression to see how many iterations it will take to either skip or fill 2015 spots... and it turns out, by convenience, that the quadratic


So, after





