*What do you notice about this problem?*

*Is it similar to another problem you’ve faced?*

*What strategies did you consider and abandon? Why?*

I’m more interested in how you think than in what the answer is.

*What do you notice about this problem?**Is it similar to another problem you’ve faced?**What strategies did you consider and abandon? Why?*I’m more interested in how you think than in what the answer is.

- PaulPaul Johnston
x+y+z=231

z-y=y-x

x=27

z=2y-x

27+y+z=231

27+y+2y-27=231

27+3y=258

3y=231

y=77

27+77+z=231

z=127

27+77+127=231

Noticed that the problem was well stated.

This problem similar to problems in high school algebra classes.

Write down the facts stated. Solve one number. Solve remaining number. Verify three numbers fit parameters stated in problem.

I noticed:

1. It's a simple algebra problem, like many I had seen in the first week of high school.

2. Since the value of one number was given, it amounts to two equations with two unknowns and therefore should be solvable.

3. I had to write it down to solve it rather than do it in my head, which is a sign of age.

x+y+z+231

z-y=y-xInteresting approach: you kept all three variables but set up an equality for later substitutions. I am so used to using only one variable whenever possible that I wasn’t sure where you were going with this. The questions for me here are, “

**Are there certain types of problems that your approach is better suited for solving? Are there problems that would be more difficult to solve with this approach?”**I’m finding it fascinating to see how different mathematicians choose to solve this problem.

- ChrisChris MacAskill
Welcome to Cake, mazed70. 👏 Great first post.

Chris, I was hoping to get @mazed70 on last weekend’s panel on “How do you make Maths Fun?”, however, a communications mixup on my end prevented that. Mazed is a maths teacher who creates wonderful maths puzzles. We’ve talked about doing another maths panel on Cake in February, topic TBD.

Here’s a recent one of Mazed’s puzzles that I enjoyed:

- PaulPaul Johnston
“Are there certain types of problems that your approach is better suited for solving? Are there problems that would be more difficult to solve with this approach?”

I do not know the answer to the above question.

The most difficult part of the solution was typing out my equations accurately. The "x+y+z+231" should have been x+y+z=231, which I have corrected. I am guessing you understood that was a typo error.

I was trained to understand what the facts were, interpret them accurately and understand exactly what is being asked. I focused on what the problem was telling me. The first three statements in my solution are the facts presented in the question. The second number(y) was related to both the third number(z) and first number(x). The value of the first number(x=27) was revealed in the problem statement.

The second sentence in the problem sets up the equality statement(z-y=y-x). This allows for the third number(z) to be defined by rearranging the variables(z=2y-x). Substituting this and the value of the first number(x=27) into the first equation of the problem(x+y+z=231) allows the second number (y) to be determined.

I chose to solve for "z" by rearranging the equality statement. "z" was the first unknown in the equality statement which is what I decided to solve for first. There was no particular reason I solved for "z" first.

However, one would think that either "z" or "y" could have been solved for given "x=27". When I defined "y" and substituted it into the 27+y+z=231, things would not work out correctly for me to determine z. Maybe you can do it and find my mistake? For "y" I came up with:

z-y=y-x

2y=x+z

y=(x+z)/2

Maybe I have made some algebra mistake or trying to do something not permitted in algebra? When going over what I did, I was going to say one could have defined either z or y and solved the problem. See if you can define y and then substitute?

For "y" I came up with:

z-y=y-x

2y=x+z

y=(x+z)/2

Maybe I have made some algebra mistake or trying to do something not permitted in algebra?Not at all.

**You are 100% correct**. It’s just that we are conditioned to working with whole numbers. Anything unusual, such as fractions or decimals, and the unconscious assumption is that we are on the wrong track. Evergreen has a great essay that touches on the challenges of our ingrained perceptions and mental shortcuts in navigating the world, and the lessons there are applicable as well to mathematical thinking. One way to deal with this ugly expression**y=(x+z)/2**is to change it into something less formidable looking

x = 27

y = (27+z)/2

**y = 13.5 + .5z**If you plug that into

x + y + z = 231

it can be simplified to

**40.5 + 1.5z = 231**Still looks ugly? Then multiply both sides by 2 to get rid of the decimals

**81 + 3z = 462**Thank you for your kind words. Here is another question in a similar style to the original.

I would use a bar model approach. What approach would anyone else take?

“Paul’s number is 38 more than mine.

Apm’s number is 24 less than mine.

The total of our 3 numbers is 137.

What are our numbers? “

I would use a bar model approach. What approach would anyone else take?

“Paul’s number is 38 more than mine.

Apm’s number is 24 less than mine.

The total of our 3 numbers is 137.

What are our numbers? “I think of the bar model as this incredibly slow and methodical method that ALWAYS gets you the right answer

**because it forces you to organize your information in a way that avoids the routine errors that come from using mental math.**And yet, I persist in using mental math.

*What do I notice?*I notice that the use of the word “mine” has the potential to add productive struggle. Would it make the problem more difficult or easier to solve if I changed your second statement to this equivalent:

My age is 24 more than apm’s.

I also notice that you’ve set the three numbers equal to 137, which is not evenly divisible by 3. My assumption is that I will end up adding or subtracting from 137 to get a number that is evenly divisible by 3. Otherwise, I’ll assume that I messed up somewhere.

**My strategies for solving your above problem**I try to use only one variable whenever possible. Since every statement is a comparison to or a fact about “mine,” it makes sense to me to set

🦆 = mine

I then set up Paul and apm’s ages in relation to 🦆

Paul = 🦆 + 38

apm = 🦆 - 24

mine = 🦆

I see 3 🦆.

I see that 38 and -24 add to 14.

3🦆 + 14 = 137

*This is where I cross my fingers and hope 3🦆 are equal to a multiple of three.*Once again, I would just use high school algebra. This time it's three equations in three unknowns.

The (slightly restated) givens are:

(1) M = P - 38

(2) A = M - 24

(3) M + P + A = 137

Substituting (1) and (2) in (3) yields

(4) (P -38) + P + (M -24) = 137

Regrouping and transposing yields

(5) 2P + M = 199

Substituting (1) in (5) yields

(6) 2P + (P - 38) = 199

Solving for P

(7) 3P = 237

(8) P = 79

Substituting in (1)

(9) M = 79 - 38 = 41

Substituting in (2)

(10) A = 41 - 24 = 17

Checking the arithmetic by substituting in (3)

(11) 41 + 79 + 17 = 137

This took way longer to type than to solve.

I did originally write it as a money problem, but changed it to numbers not in context to avoid whether I should use pounds or dollars.

The fact that the total is not divisible by 3 was not a deliberate choice.

The wording that apm’s number is less than mine was deliberate, rather than saying my number is 24 more, just because I thought it was a bit more complex/ interesting that way.

My solution would be to draw three bars and compare what I know.

It is interesting to see different approaches to the same problem.

- PaulPaul Johnston
Thank you for taking the time to show that the problem could be solved by defining "y" !

Your examples "reset" or "cleared" my mind so that I could go back an find a multiplication error in my calculations in substituting "y" that made "z" come out to be some weird decimal value. Before I made my last post, I went over my work several times and glossed over the same multiplication error and did not catch it. Your showing me what I thought was correct, that the problem could be solved by either defining "y" or "z", cleared my mind to immediately spot the multiplication error. Thanks again!

You were interested in the various approaches people came up with in solving the same problem. As I was trying to find the error in my defining "y", I asked my son what was I doing wrong. He first said he just wanted to work the problem. I gave him the three statements that defined the problem:

x+y+z=231

z-y=y-x

x=27

I verbally described the problem. He rearranged the above and performed row matrix operations and came up with the answer to "z" and "y".

Glad I could be of help with my walkthrough. Explaining a solution also helps to crystallize my thinking on why a particular approach is more likely to achieve desired results.

Kudos to your son on his approach. Back when I was in high school, they did not teach about matrices in math class. I believe it’s only in the last twenty years that matrix operations were included in the Algebra 2 curriculum. So when I think of the right tool to use, it’s unlikely to be the first one I grab. But like @mazed70’s bar method, it’s a strong organizing method of the data you’ve been provided, reducing the risk of errors.