I caught this article in my feed this morning and it got me thinking. It discusses a simple problem:

Take the numbers 1 to 100, square them, and add all the even numbers while subtracting the odd ones!

The article itself lays out 7 ways to solve the puzzle. One is quite boring (just type everything in by hand like a calculator). I think the point of the article is cool - specifically, that there are often many ways to solve a problem. It did make me think about how I would work with a student to solve this problem though. I feel like the article is focused on experienced R users (note the general lack of comments in the code) rather than to a new user.

I think my favorite solution is somewhat similar to the first solution that he immediately thought of…well, the one right after the boring one…with a few extra steps. I can imagine a dialog with a student looking like this…(Note, I’m a physics professor by trade, so when I’m teaching students to code, I’m often pointing them in the direction of useful functions so that they can solve the physics)

Solving the problem

Me: Ok, so what do we need to solve this problem?

Student: Well, it would be really nice if we had all of the even numbers in one vector and all of the odd numbers in another vector.

Me: Great! Let’s check out the seq() function. Put these lines into your interpreter and see what pops out:

seq(from=1,to=10,by=2)
## [1] 1 3 5 7 9
seq(from=2,to=10,by=2)
## [1]  2  4  6  8 10

Can we use these to generate what you want?

Student: Yes! I think I can do this:

odds = seq(from=1,to=100,by=2)
evens = seq(from=2,to=100,by=2)

Ok, now can I just square these things?

Me: Whenever I’m unsure of something, I like to try it out on a shorter vector that I can check by hand. Like this:

x = 1:5
x^2
## [1]  1  4  9 16 25

Does, \(x^2\) produce what you expect?

Student: Yes, I think it does! From here, I think I can just square, sum, and then find the difference. I’ll try that out:

sum(evens^2) - sum(odds^2)
## [1] 5050

I get 5050, does that make sense?

Me: That’s what I have too.

Challenge accepted

I’ll end by noting that the author points out that the answer to this problem is identical to “What’s the sum of all numbers 1 to 100?” I thought that I’d prove that for fun.

So, we are looking for the sum of all the even numbers squared between 1 and 100, subtracted by the sum of all the odd numbers squared on the same interval. We can write the solution \(S\) as:

\[S = \sum_{n=1}^{50} (2n)^2 - \sum_{n=1}^{50} (2n-1)^2 = \sum_{n=1}^{50} \left[(2n)^2 - (2n-1)^2\right]\]

The term in the square brackets matches the form for the difference of squares. Let’s use that to re-write:

\[S = \sum_{n=1}^{50} \left[\left(2n - (2n-1)\right)\left(2n + (2n-1)\right) \right]\]

This simplifies to:

\[S = \sum_{n=1}^{50} \left[\left(1\right)\left(2n + (2n-1)\right) \right]\]

The 1 can just go away, and let’s split the sum apart again:

\[S = \sum_{n=1}^{50} 2n + \sum_{n=1}^{50} (2n-1)\]

The first term is just the sum of all even numbers between 1 and 100, and the second term is the sum of all odd numbers between 1 and 100. Therefore:

\[S = \sum_{n=1}^{100} n = \frac{100(100+1)}{2} = 5050\]

I should note that this result requires that the maximum term must be even. For example, the sum of all numbers from 1 to 99 is 4950. But, the same question (evens^2 - odds^2) for 1:99 is somewhat different:

sum(seq(2,99,2)^2) - sum(seq(1,99,2)^2)
## [1] -4950

To re-cast the proof for an odd max value, I’d have to include zero as a term for the evens:

\[S' = \sum_{n=0}^{49} (2n)^2 - \sum_{n=0}^{49} (2n+1)^2 = \sum_{n=0}^{49} \left[(2n)^2 - (2n+1)^2\right]\]

And note that the left hand term simplifies to -1…

\[S' = \sum_{n=0}^{49} \left[\left(-1\right)\left(2n + (2n+1)\right) \right]\]

So, for an odd max value, the answer is just -1 times the sum of all values up to that max value.

\[S' = - \sum_{n=0}^{99} n = - \frac{99(99+1)}{2} = -4950\]

This was just too cool not to write up.