The problem of "penny piles" was presented on March 13th, on account of half the class being brain-dead from taking the CSC148 test.
The Problem: There are two drawers of pennies, and the basic premise involves taking half the pennies in one drawer and putting it in the other (if number of pennies is divisible by two) to make any combination of pennies. The question is if every combination is possible, and what happens when you start with different amount of pennies.
The Plan: Despite not having taken the exam that morning, neither my partner and I were really up for much thinking, so our plan was basically the same as for the other problem solving exercises: do random things until we find a pattern, and then think about why the pattern works.
In class, we tried working backwards and semi-systematically found almost every combination of numbers (we got lazy and did not actually finish the chart containing numbers). What we could have done instead that would have been very similar was draw a tree diagram of the possible results of the original problem of 64 pennies in the left drawer and none in the right drawer, which I have painstakingly recreated here:
(The right half of this tree repeats itself but in different order so I have omitted a large part of it.)
This tree diagram shows that it is possible, if we include the amount in both drawers, to make every number between 0 and 64 if we start with 64 and 0 in the left and right drawers respectively. This is likely the least elegant way to prove such a thing.
Looking back: After having carried out the plan, it worked reasonably well, but our initial thinking should have been more organized and we should have had a tree structure to begin with.
The Problem: There are two drawers of pennies, and the basic premise involves taking half the pennies in one drawer and putting it in the other (if number of pennies is divisible by two) to make any combination of pennies. The question is if every combination is possible, and what happens when you start with different amount of pennies.
The Plan: Despite not having taken the exam that morning, neither my partner and I were really up for much thinking, so our plan was basically the same as for the other problem solving exercises: do random things until we find a pattern, and then think about why the pattern works.
In class, we tried working backwards and semi-systematically found almost every combination of numbers (we got lazy and did not actually finish the chart containing numbers). What we could have done instead that would have been very similar was draw a tree diagram of the possible results of the original problem of 64 pennies in the left drawer and none in the right drawer, which I have painstakingly recreated here:
(The right half of this tree repeats itself but in different order so I have omitted a large part of it.)
This tree diagram shows that it is possible, if we include the amount in both drawers, to make every number between 0 and 64 if we start with 64 and 0 in the left and right drawers respectively. This is likely the least elegant way to prove such a thing.
Looking back: After having carried out the plan, it worked reasonably well, but our initial thinking should have been more organized and we should have had a tree structure to begin with.
