Pi Club: Round 2

First things first, Pi Club has a new member – and SHE is a wonderful addition to the group! WOOP WOOP, girl power!

Today, we worked on two problems. First, the lightbulb problem (found here):

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person.

Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6…). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9…). This continues until all 100 people have passed through the room.

What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

They didn’t take too long to work this one out. Great conversations were had.

The second problem we worked on was the (modified) chessboard problem (also found here):

How many rectangles are there on an 8×8 chessboard?

(They had already seen and solved the problem of how many squares there are on a chessboard.) Their work here was particularly interesting to watch. They ended up splitting up into two groups, each breaking out of the confines of their small individual whiteboards to work on the big white boards at the front of the room. (Putting the VNPS to work!)

Pi Club 11-27 1
Solving a simpler problem

The two groups ended up finding two different solutions to the problem (neither of which I had anticipated – the students’ creativity never fails to amaze me!), but one of them really blew me away. Here’s how it goes:

  • We know that the solution for the number of squares is 1×1 + 2×2 + 3×3 + … + 8×8. It therefore stands to reason [student’s exact words] that the solution for the number of rectangles is 8×1 + 8×2 + 8×3 + … + 8×8 + 7×1 + 7×2 + … + 7×8 + … + 1×1 + 1×2 + … + 1×8. [This can be summarized in a table, but this wasn’t part of the student’s solution.]
  • Note that 8×1 + 8×2 + 8×3 + … + 8×8 = 8(1 + 2 + 3 + … + 8) = 8(36).
  • Therefore, the solution is 1(36) + 2(36) + 3(36) + … + 8(36) = 36(1 + 2 + 3 … + 8) = 1296.

IMG_20151127_124432751_censored

Isn’t that just BEAUTIFUL?!

Pi Club 11-27 3
VICTORY DANCE!

I love this job.


 

P.S. If you have any interesting problems I could add to my arsenal, send them my way!

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2 Replies to “Pi Club: Round 2”

  1. I like making the board irregular (say 8 x 12) and ask the same question of ‘how many rectangles’. The elegance that often occurs is recording the data in a table that is in the same shape as the board. Students quickly realize that triangular numbers are appearing along the first row and column and the rest of the table is a product of the cross and down triangular numbers. So the 8 x 8 checkerboard would be 8(8+1)/2 x 8(8+1)/2 = 36 x 36 = 1296 rectangles. The 8 by 12 would then be 8(8+1)/2 x 12(12+1)/2 = 36 x 78 = 2808 rectangles.

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