(blog): Happy Consecutive Number Day!

It's seems like every day, like every number, has something special about it, but as George Orwell might've said, but wisely chose not to, "Some dates are more special than others". (Had he said something about some numbers being more equal than others, that could've been a problem, but would've fit well into his theme. But that's a subject for another blog post.)

Today is Consecutive Number Day because it is 12/13/14. We've had a Consecutive Number Day for each of the past 11 years -- TWO, if you use the Commutative Property of Dates (also known as the DD-MM-YY format). This last fact makes this year's date even more special because there won't be a 12-13-14 is London or Madrid or pretty much everywhere else in the world where they use that format because there isn't a 13th month.

Finally, it's special because after 12 years in a row, this one will be the last one for a long time. Now, don't fret yourself into a worry. You don't have to wait until 2103 for another one ... unless you're some kind of crazed purist. Me? I like to run the numbers, have fun with the numbers.

The first Consecutive Number Day that I took note of happened when I was in ... kindergarten, no, wait, Pre-K! Yeah! That's the ticket! ... in 1978. It was 5/6/78 and we even had to pause at 12:34 pm (I was asleep for 12:34 am) to note that it was 12:34 5/6/78.

Eleven years later, digital watches were common and we knew the correct time to the second, if we ever bothered to set it correctly, so we knew when it was 1:23:45pm 6/7/89. We could do the same the following year, although 7/8/90 wasn't quite as much fun.

Two thoughts come out of this. First, given the accuracy of today's computing, I had to wonder if the atomic clock could tell us when it was exactly 1:23:34567891011 12/13/14 (and, if so, did some geek take a selfie of it?). Second, it gives us something closer to look forward to.

Hold the date now in your datebook app for 1/2/34. Sadly, there isn't a 5:67 am or pm. Not even in London. Or even in Pasadena, where they'll be hosting the Rose Bowl because New Year's Day is on a Sunday.

Article: The Sums of Consecutive Squares

Fun fact for the 365th day of the year:

10² + 11² + 12² = 365 and 13² + 14² = 365

So 365 is the sum of two sets of consecutive squares, but, more importantly, those two sets are themselves consecutive: {10, 11, 12} and {13, 14}.

Now that's interesting! Okay, so it's also a co-incidence, really. Sums of consecutive squares have to add up to something, and, occasionally, those "somethings" will be the same number. But can we write a general rule for this?

Of course, we can. If we can write it, we can (hopefully) solve it. (Again, of course I can solve it, or I wouldn't be asking the question, but I'm sure that there are many, many rules which we could pose which I, personally, couldn't solve. However, this is a simple one to work with.)

At the very simplest level, we have the sum of the squares of two consecutive positive integers. It's obvious that these two numbers can't equal the sum of two higher integers, so the sum has to be of only one number. Not much of a sum, I grant you, but we're starting with a trivial case.

We want to find consecutive positive integers, a, b, and c such that

a² + b² = c²
so we'll use the variable n to stand in for the lowest integer, (n+1) for the next consecutive integer, and (n+2) for the third consecutive integer. Now we can rewrite the equation as

n² + (n+1)² = (n+2)²

Squaring the binomials, we get:

n² + n² + 2n + 1 = n² + 4n + 4

Combing like terms gives us:

2n² + 2n + 1 = n² + 4n + 4

Rewrite this as a quadratic equation by subtracting the right side of the equation from both sides:

n² - 2n - 3 = 0

Which factors into: (n - 3)(n + 1) = 0. Therefore, n = 3 or n = -1, but because we want a positive whole number, we'll discard the -1 and accept the 3. That makes the three consecutive integers 3, 4, 5 and, therefore, 3² + 4² = 5² .

But, of course, you already knew that. So why do all that work? Because now we can move up to four or five consecutive numbers. We can use the same procedure to find solutions to

a² + b² + c² = d² or a² + b² + c² = d² + e²

To save space, and to be as annoying as those textbook writers of my youth, I'll leave a, b, c, d to you to try. I'll give you a hint: there aren't any positive integer solutions, but you can prove that for yourself instead of taking my word for it. Go ahead -- challenge authority!

For the sum of the squares of three consecutive positive integers equal to the sum of the squares of the next two integers, this is the equation we write:

n² + (n+1)² + (n+2)² = (n+3)² + (n+4)²

When the dust settles, what will be the value of n? If you didn't get it, you weren't paying attention. It's in the first paragraph of this article. The solutions are n = 10 and n = -2. Once again, we toss the negative and we're left with: {10, 11, 12} and {13, 14}.

This brings two closing questions: The obvious question is what seven consecutive positive integers a, b, c, d, e, f, g fall into this pattern? (I didn't say that the answer was obvious, but it's easy to figure out.)

And another question about another pattern: In the first case, we threw out the solution n = -1. In the second case, we discarded n = -2. I'll go ahead and tell you that in finding the answer to the next sequence of numbers, you'll have to get rid of the solution n = -3. My question: will the negative solution we discard always have the same absolute value as the number of terms on the right side of the equation?

I'll leave that as an exercise to the reader. You have a whole, exciting, brand New Year to work it out!

Sums of Squares of Consecutive Numbers

Continuing what I started on Tuesday's post:

Giving the Pythagorean Theorem, a² + b² = c², find three consecutive numbers for a, b and c.

The answer, of course, is everyone's favorite Pythagorean Triple, 3, 4 and 5.

But suppose instead of three consecutive numbers, we had five consecutive numbers, which were split with the three smaller values on the left and the two greater values on the right? That would give us:

a² + b² + c² = d² + e²

The answer is in the previous post. I'll omit it here in case you want to work it out.

Now, a couple approaches could work here. The first, and probably best if you plan on going further, is to replace the variables with n, n+1, n+2, etc., and then using FOIL (or a "FOIL"-free alternative if you hate "FOIL"), combining like terms and solving the resulting equation.

The other, which I can use on Day 1, is guess and check. Okay, stop laughing and rolling your eyes, and hear me out.

First of all, most of my students haven't handled a scientific calculator very much let alone a TI-83, 84, or N-Spire. (Yes, we had a bunch donated to the school as part of a technology initiative. Unfortunately, they aren't allowed to be used during the Regents exams, so we have to switch back to the older calculators. But that's a rant for another day.) An activity like this could be a simple and thoughtful first exercise.

Second, many of them have little or no Number Sense or Estimating skills. How would they approach the problem? Would they try 3, 4, 5, 6, 7 first? When that doesn't work, will they move to 4, 5, 6, 7, 8, or will they jump a little higher a little faster? But if they don't try every combination, how will they know if they went too far and passed the answer?

Then, after finding the answer and comparing it to 3, 4, 5, what would their first guess be for
a² + b² + c² + d² = e² + f² + g² ?

Naturally, all of this occurs to me during the last week of classes, after the
final exam has been given.