Some random numbers were put in the bricks on the bottom row and they were then asked how are all the other numbers generated? I stood back and let them argue for a couple of minutes before they figured that the numbers from two bricks add to give the number on the brick above.
"Can you predict what the top brick will be if the bottom bricks are all 1s?" I asked.
A brief but furious debate ended with "The second row will be all 2s, the third row will be all 4s, the next row will be all...8s so the top row will be 16! It will be 16!"
"Prove it," I said which they duly did to hearty cheers. I followed up with asking for their predictions for all 2s on the bottom row then all 10s and again they were spot on with their predictions.
So I decided to ad lib a bit:
"Right, your target is to get 1000 in the top brick but you must have the same number in all the bottom bricks."After a flurry of trial and error (or trial, feedback and refining to be exact) they got to the point where 63s on the bottom row gave 1008 in the top brick.
"Too big! It's too big! Try 62!"
And then the following happened:
The students loved it and I consolidated by giving them this question from the Problem of the Week section on the University of Waterloo's CEMC Problem of the Week site..