# Problem: February Problem to discuss

I am going to push everyone out of their comfort zone, read the problem below and solve it in two different ways.  One way can use a formula or the common algorithm but you need to post a different way to think, too.  Let's support each other in this endeavor.

The perimeter of Square $A$ is $28$ feet less than the square $B$ and the area of the square $A$ is $161$ less than the square $B$ if the length of the square $A$ and $B$ is $x$ and $y$ respectively.

What is $x+y$?

Source JMM ### First Way, Brute Force

I like the challenge of solving two different ways. I spent a few moments thinking about how I could do it, and deliberately avoided using too much equation-solving, as I planned to make that my second method. I decided to try guess-and-check, but in a last moment of lazy inspiration I realized I could have Excel do the guessing for me. I threw together this little table: Since the perimeters had a difference of 28, I knew the difference between side-lengths must be 7 because 7*4 = 28 so:

I listed possible values for x,

added 7 to each to get a possible y,

found the area of both squares

and then subtracted to find the difference.

Let me know if the table is unclear.

I then found the value of x that results in an area difference of 161. Turns out that's x=8 and y=15, so x+y=22.

I'll try to write some equations to solve next... ### Changed Direction, kinda...

In reply to by Patrick Rodenborn

So I started solving the equations, and as I neared the end I saw something interesting and got distracted from finishing the solving.

First, my two equations:

x+7=y

y^2 - x^2=161

Then substitute the first equation in for y to have:

(x+7)^2 - x^2 = 161

Distribute to get:

x^2 + 14x + 49 - x^2 = 161

Here I noticed that the x^2 will cancel, and wondered why and how I could have used that info earlier. I pictured one square, then another square growing from it by 7 in two directions to make the bigger square, and noticed that I was imagining an area model. I then drew the following: Since I know the difference in areas is 161, I want to just focus on the difference part. There's a rectangle that is 7*x, another that is x*7, and a square that is 7*7. All of those should add to make 161.

Solving for x gives 8, so y is 15, so x+y=22.

I'm looking forward to seeing what others do!

Patrick ### I went to algebra

In reply to by Patrick Rodenborn

I also figured out okay, 7 different... so x squared - (x-7) squared is 161.   I'd say this is an example of "this is why formulas are a good thing," because it's easy to get to x being 15 that way. I thought about how 161 is 7 x 23 but didn't get anywhere... ### solution with a table

Hi all,

I started thinking about how the difference of squares result in the odd numbers: 4 - 1 = 3, 9 - 4 = 5, 16 - 9 = 7, etc. and thought maybe the area of square B minus the area of square A would be between two consecutive squares. What two squares would give me a difference of 161? I played around a bit and found out that:

812 - 802 = 161

So, the side length of square B is 81 and the side length of square A is 80? The problem is that the difference in the perimeters of these two squares is only 4:

4(81) - 4(80) = 4

So, I guess the squares aren't consecutive. My next attempt was with a table. By the way, is the question What is ? meaningful? Why not just ask for the side lengths of the two squares? Or what are x and y?

Looking forward to seeing other solutions,
Eric ### x + y

I wonder if using algebra to solve for x+y instead of separately finding x and y does anything interesting? Is it easier or more challenging? ### Consecutive odd numbers

How can the difference between the perimeters of two squares be 28? Since all four sides of a square are the same length, the 28 extra units of perimeter must be evenly distributed to all four sides. In other words, the sides of the larger square must each be 7 more than the sides of the smaller square.

What about the difference in the areas? I know that when I build out a square to a bigger square by adding to the side lengths one unit at a time, I am adding consecutive odd numbers, so I need a string of 7 consecutive odd numbers that adds up to 161 (because it will take an increase of 7 to get from x to y). If I divide 161 into 7 equal pieces, I get 23. That will be my middle number. Stepping up and down from 23 by twos gets me this set of numbers: 17 + 19 + 21 + 23 + 25 + 27 + 29 = 161. The first amount of area that I add on to my smaller square, then, is 17. Here's a picture that shows how adding one to the sides of a square increases its area by an odd number and how that number is related to the original side length: So the 17 is the red area in my picture, which is twice x and one more. In other words, x must be 8.

The side length of the larger square (y) is 7 more than x, so it is 15.

The sum of x and y is 23. (Interesting that that number had already come up in my calculations... is there significance to that?)

Here's another way I thought about it. One of the pieces of information in the problem is literally telling us about a difference of two squares (as Eric noted). Because I know the factoring pattern well, I see that y2 - x2 = 161 means that (y + x)(y - x) = 161. I looked for factors of 161 and found 7 and 23. I found the point exactly in the middle of 7 and 23, which is 15. That number is 8 away from 7 and 8 away from 23. So I can write my factored equation as (15 + 8)(15 - 8) = 161. In this case, I actually did have the sum of x and y before finding the individual values, but I wouldn't have relied on it because I wanted to make sure the two numbers differed by 7 as well. (I could have settled on 1 and 161 as factors of 161 and ended up with (81 + 80)(81 - 80) as my two factors and those would not have satisfied the perimeter constraint.

-Sarah ### One of our courses now

In reply to by Sarah Lonberg-Lew

One of our courses now includes using division to find the middle number in strings of consecutive numbers ... but it annoys me because they never get to why you'd do that so to the students it's just another thing to put in the calculator (unless, of course, you're in the tutoring lab and I get you there ;))      Here, I got to "okay, 7 x 23=161"  and was thinking about the "consecutive odd number" pattern, since yesterday our math literacy class plotted rectangles w/ same perimeter and integer lengths and widths and we saw it there...   but I didn't make the rest of the connection.