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Mission Possible: The Prison Problem

Link: The Prison Problem

This problem didn’t take me too long to figure out. I knew that I had seen it before (though in a different context) so I had hints of strategies floating around my brain. Before going into any math though, I thought about what conditions would leave a cell open and would conditions would leave it closed. Everything was opened first. Then, multiples of 2 would have closed them, and so on with other multiples. So, if a number has an even number of factors, it would end up closed and if it has an odd number of factors, it would end up open. This gave me a direction with which to carry out my trials. All I needed to do was figure out how many factors each number between 1 and 100 had!

As always I started off by organizing my information and doing a trial run of some numbers in hopes to identify a pattern. The idea of calculating how many factors all the numbers between 1 and 100 had would be way too time consuming so I sought out an easier and more mathematical route. I tried out the escapist’s strategy on what would be the first 20 prison cells to give me a starting point. Once I had numbers I was able to confirm my theory that an odd number of factors leaves it open and an even number of factors left it closed. There was a hint in my brain somewhere that was telling me that prime factorization would help out with this problem, so I also went about writing out the prime factorizations of these as well. I couldn’t see the connection right away, but I had remembered something of a theorem from my Elementary Number Theory class a few years ago that would tell me how many factors a number has. I looked this up and that was the last bit I needed before I was able to clearly see how to finish off this problem. The theorem states that the number of factors of a number is the product of the (exponents+1) in its prime factorization. So the next question I asked myself was: how can I use this information to create an odd number? Below is my progress of thinking to reach my conclusion:

  • How can a product equal an odd number? It must have no even numbers in it
  • The product is determined by the exponents+1, so my exponents can ONLY be even numbers
  • SO! All perfect squares are open
  • Any other case can be rewritten as the product of perfect squares, because my exponents are only even numbers
  • The product of perfect squares is once again a perfect square so I can conclude that only situation in which a prison cell is open is if it’s cell number is a perfect square!

Cow, Pig, and Sheep Problem

Link: Cow, Pig, and Sheep Problem

When I first read through this problem I thought it was so easy. I completed it so quickly and wondered even why we had been assigned this. Then I realized that I didn’t read the whole thing and that I needed to have 100 animals. So then I went about solving the ACTUAL problem. I started off the way I start off any problem: write down what I know. Doing this led to two equations and three variables, which I knew I couldn’t solve. There had to be something else that I could extract from the problem. Combing through it again, this time much more frustrated, I decided to try and take the constraint on having at least one of each animal and turn that into something that I could use. All that resulted in, though, was my changing my original equations so that at least one of each animal would be accounted for. The only way I could think to solve this was to start substituting guesses in for how many sheep there were, thus reducing my two equations to two variables and therefore making them solvable. I started off by guessing there were 51 sheep. I knew that I needed an odd number of sheep because I had already accounted for one and I needed to bring the cost to $100 even. This gave me a weird answer that didn’t work within the constraints of the situation so I moved on and tried 85 sheep. This didn’t work either. I had the feeling that there was something I was missing, so I took a break and went back to the problem a few hours later. That break was exactly what I needed. After looking at the algebra that I had used to solve my previous attempts I saw that I could generalize this situation to help me make better guesses for the number of sheep. My previous solutions kept giving me negative answers, so I needed to rework my guesses so that I could produce positive answers. The cost of my pigs, for example, had to be less than the remaining total after I purchased all my sheep (and one of each animal). So that meant that 3(97-s)<86.5-0.5s. If I solved this then I could find a restriction on s, thus narrowing down the possibilities for me to begin my guessing. This led me to the fact that I had to have more than 81.8 sheep, so I began the remainder of my tests there. After some time I found my solution. It was annoying to solve this through an educated guess and check method, and I wish I could have remembered a more efficient way to do so, but sometimes math is all about the long, tedious, number crunching!

Coke and Sprite Problem

Link: Coke and Sprite Problem

This was an interesting problem to work on. I only watched the video twice, and both times I spent pausing and staring to see how much liquid was being transferred in that eye dropper. Right away, I knew that would be my key to solving this problem. Every time I watched the video though, I failed at figuring it out. How frustrating! I knew exactly how I wanted to solve this problem, and yet I had a key numerical value missing and therefore couldn’t really go anywhere. So I started googling how much the average eye dropper would hold.. I finally agreed upon the fact that they hold anywhere between 5 and 15ml so I knew I had a few calculations ahead of me. Now I had everything I needed to know to solve the problem! Being the avid pop drinker that I am I knew that each can contained 355ml. My method was simple: assuming the eyedropper held x ml of pop, I would do a step by step percentage calculation of the original pop and the new one, based on the new total quantity of pop and how much of each were mixed in. My instincts told me that it would probably end up being the same (assuming we transferred the same amount both times) but another part of me was dead set on the fact that that was way too obvious of a solution and that couldn’t be true. Another part of me was tempted to try it out and taste them to confirm my results, haha. In the end I did the calculations a ton of times with the eyedropper holding 5 ml, 10 ml, 11 ml, 12 ml, 13 ml, 14 ml, and finally 15 ml. They all gave me the same answer: the percentage of original pop was identical in both of them. I redid all the calculations with more decimal points because I just didn’t want to believe that they had equal concentrations! In fact I still don’t want to believe that they have different concentrations, but I have no clue how to prove that! So while the math is right in front of me, I still find myself pondering other ways to do this problem in hopes of getting a different answer, but I still can’t think of anything! Frustrating!!

Salesman Problem

Link: Salesman Problem

When I first read this problem, my gut instinct was to claim that it was impossible and that there was no way to solve this problem. Instead, I reassured myself that I love solving problems like this and began to work through it using my fail-safe method of breaking it down into smaller pieces, writing down what I know, and writing down what I want to know. Once I had translated the problem into a format that I was more comfortable looking at, the solution came to me pretty quickly. I had a little fun with this problem and pictured myself in the shoes of the salesman.

Here I am standing at this woman’s door and she tells me that the product of her children’s ages is 36. So this tells me that step one is to write down all possible combinations of 3 numbers that multiply to 36. Awesome. Once that’s done we go into the crazy information that includes no numbers. Time to pull out the logic!

So I’m standing at the door and the woman tells me that the sum of her children’s ages add up to the house number next door. Here is where I got stuck for a while. Obviously I then took all of my combinations and wrote down their sums. But I’m not at the house, so I have no clue what number I’m looking for and suddenly I have 7 different possibilities. Is is supposed to be an odd number? If we’re on the same side of the street as this woman’s house, and her house number is an odd number then my sum should be? Or should it be an even number? I pondered these for a minute, but they lead me nowhere. So instead I stopped and listened to the words that were being exchanged around the room as others were working on the problem. This actually got me nowhere. I found no inspiration, only frustration! I stared at the paper for a little while longer, reared the problem a few times and then it hit me. He needed more information! There was a third clue! Therefore, if I were to put myself back into the salesman’s shoes, I know that I had to have more than one answer that gave me the house number next door. Otherwise I would have known right away and the problem would have been solved! So it seemed as though the key to this problem was to stay in the salesman’s shoes and experience the problem firsthand. So I had 4 sets of ages that produced repeated results. Two gave me the sum of 13 and two gave me the sum of 16.

The woman’s next clue was that her oldest plays piano. A brief moment of frustration again as I wonder what in the world this could possibly do with getting the solution to this! So I looked at my groups of ages: I had a 1, 3, 12 and a 1, 6, 6, and a 2, 2, 9 and a 3, 4, 9. My husband is a piano player so my mind instead starting thinking about what ages kids usually are when they’re playing the piano and yet all the oldest children in these situations could very easily be playing piano and so that thought lead me nowhere. I thought about this for a long time until I realized my one key point that lead me to my solution: this clue leads to an obvious solution, assuming that I new that the house number was 16 or 13. So let’s think about this in that way. Suppose the house number was 16. I then have the two possibilities of children with ages 1, 3, 12 or 3, 4, 9. In either of those cases, being told that the eldest plays piano leads to no logical conclusion. If the house number was 13 then I would have the two possibilities of 1, 6, 6 or 2, 2, 9. If I was told that the OLDEST child is playing piano, than that means that there is a singular oldest child. No oldest twins! So the 1, 6, 6, has been ruled out and my solution is 2, 2, 9! Woohoo! Joy!

I felt pretty accomplished when I solved this one!!

Marriage Problem

Link: Marriage problem

When I first read this problem, I thought that I knew exactly what to do. It had me thinking back to teaching fraction multiplication word problems to my grade eight students and telling them that of means multiply. So initially I thought that when 3/5 of the women married 2/3 of the men I wanted to multiply those fractions together. Boy was I ever wrong! I knew immediately that this was too easy of a solution and that I had to have been wrong. I then began my problem solving steps that I use every time I’m given a math word problem:

Step 1: Write down what I know
Step 2: Write down what I want to know
Step 3: Create a visual
Step 4: Make connections between what I know and what I want to know
Step 5: Perform any calculations
Step 6: State my answer
Step 7: Check my answer

Once I began thinking about what I knew from this problem I was able to write an algebraic equation to model the situation. If 3/5 of the women are married to 2/3 of the men, then that means that 3/5 of the women has to equal 2/3 of the men. This equality let me solve for one variable so that I can use it in a substitution when I have my next equation. When I wrote down what I wanted to know, I was easily able to create an expression that would give me the fraction of the population that’s married. It had to be the number of married people over the total population. From there I could substitute in from my equality and simplify, thus giving me my final answer of 12/19. While writing these steps out may seem simple, it took me quite a while to do.

To be honest, I kept trying to create another equation, thinking that the equality wouldn’t help me simplify my fraction at all. I played with so many different numbers and even invented a town with a population of women that I chose and went from there, in hopes that some solid numbers and data would help me figure out what to do. I got really frustrated because I knew that I had seen this question before, many times during my mathematical career. I had figured it out those times, why couldn’t I figure it out now? I was over-thinking. I am always over-thinking. What would my brain do if it wasn’t thinking? I don’t let it rest, I let it run into overdrive instead. I took a break and re-approached the question a few hours later and with a fresh mind I was able to play around with the algebra and set myself on the right path. I had reached out to some friends of mine that were always good at math and had them check over my ideas for me. I figured out what direction to go into before they responded, though.

When I think about how students would feel performing this problem, I am torn. Would they apply the same steps that I did? Has anyone ever actually taught them how to go about solving word problems? The frustration that I felt while trying this problem would probably be 1o times worse if I were a high school student. It really showed me the importance that I must put into my practice on how to solve math problems instead of simply just solving them. I was able to reach the solution with continued critical thinking; would my students be able to do the same thing?