Friday, January 1, 2010

HAPPY 2 x 3 x 5 x 67! Let The "Problems" Begin!

May this new year and decade bring happiness and prosperity to each of you now that the 2KO's have come to an end! 

BTW, the italicized symbol in red is my submission for the name we should give to the past 10 years. What do you think of it? Let me know if you came up with one of your own. According to Time Magazine, no one has yet created a name which has caught on (and dozens were listed!).  Also, I will avoid debating those who strongly believe that the first decade of the 21st century ends a year from now!

As MathNotations begins its 4th year, it has become an annual tradition for math ed blogs to challenge their readers to discover interesting facts about the number symbol representing the new year, in this case, 2010, or Twenty-Ten, for those who are as committed to multiple representations as I am!

Those who know me can anticipate that I would recommend making this an exercise for our middle schoolers. Here are a couple of ideas:

"In your group, list as many observations as you can about the number, 2010. Your team's score will be based on both quality and quantity. For example, an observation like "2010 is even" would only earn 1 pt, whereas "2010 must be divisible by 3 because the sum of its digits is divisible by 3" would earn 2 or 3  points since it contains both a fact and an explanation."

Another idea might be to have students write interesting word/number problems involving 2010 for the class to solve. Of course, to obtain credit the student posing the questions must  provide correct answers and solutions!

Your turn...

A final note ---

Some of you may have noticed that I've enabled Comment Moderation due to the number of spam comments which have gotten through. I held out for as long as I could. I do check throughout the day, so, hopefully, this should not prove problematic for my readers.


Eric Jablow said...

Perhaps last year's Putnam Competition will inspire you. It had three '2009' problems.

Dave Marain said...

Thanks, Eric, and Happy Twenty-Ten!

I believe I sat for the Putnam back in 1965 and it was a humbling experience. I may have gotten one or two right (or parts thereof) but, like any great math contest, I learned a great deal of math by going back over the questions and discussing them with my friends and professors. In retrospect, similar to the MAA Contest at that time (it wasn't called AMC then), I might have done better had I actually practiced with old contests for several weeks. Preparation does help...

Anyway, I will look at your link and be humbled all over again!

Mike said...

2KO's is a great name for the last decade. Good blog.

Dave Marain said...

Thanks, Mike, and Happy Twenty-Ten!

David said...

2010 is the sum of five consecutive squares.

2010 is the product of the 19th prime number and the 19th composite number.

Dave Marain said...

From one David to another--
Very nice observations! Give us a clue how you came up with these.

Lagrange proved that every positive integer can be expressed as the sum of 4 or fewer squares! In fact, 2010 = 44^2 + 7^2 + 5^2 and I'm pretty sure you could find other representations. Since 5^2 = 3^2 + 4^2, this leads to writing 2010 as a sum of 4 squares. So what is your result or is that our challenge problem!

The 19th composite observation surprised me so I felt the need to check it.
Let's see...
I get that 30 is the 20th composite (pos integer), but please check my work.

Dave Marain said...

I'm guessing that you would regard 1 as neither prime nor composite so that your observation would work! Sorry 'bout that!

mathmom said...

That's generally the way it's taught in middle school -- that 1 is neither prime nor composite.

Cool thread :)

Happy New Year!

Dave Marain said...

I would hope every textbook K-12 defines 1 that way. That was just my oversight. I was probably thinking of the term "non-prime" rather than "composite". 1 is certainly the former but not the latter!

And, yes, this thread is leading to a nice discussion. I was hoping you might comment on the previous two posts which are much more important to me:
"How much quadratic trinomial factoring in Algebra I?"
"The 12x^2 + bx + 15 investigation"

mathmom said...

I don't know anything about the standards, etc. so it's hard for me to comment on what's a good use of classroom time in a high school class. I do think that factoring is a good "puzzle" or problem-solving activity that helps grow good problem solvers. I'm not sure if it's "still" a core basic skill that kids need to have, given the ubiquity of tools to do the work for one. The ability to do the symbolic manipulation never hurts, especially for those going further with mathematics, but in terms of something every high school student should learn before they graduate, I'm not passionate about it.

I liked your challenge activity, though I think it is too hard for most Algebra I students to do on their own.

I used a couple of your problems (palindromes and the trillion multiples thing) with my oldest group before the break -- hopefully I'll get around to writing up some observations sometime soon :)

David said...

Hi Dave,

My strategy for finding these representations was rather mundane. I simply searched the OEIS for 2010, and scrolled through the hundreds of matches to find the few that were interesting.

I will leave the problem of finding five consecutive squares whose sum is 2010 as a challenge. It's not very difficult.

Dave Marain said...

Thanks for clarifying that. I definitely need to be more alert when I read your comments. I missed the "consecutive" part when I originally read it!

The algebra as you said is straightforward but you made me think about asking students to describe all numbers which can be expressed as the sum of the squares of 5 consecutive integers. This could start out as a prealgebra investigation for middle schoolers in which they collect some data:

(-2)^2 +(-1)^2+0^2+1^2+2^2 = 10
(-1)^2+0^2+1^2+2^2+3^2 = 15
0^2+1^2+2^2+3^2+4^2 = 30

It would be obvious without algebra that the sums would be divisible by 5 but the other factor is less obvious:
10 = 5 x 2 = 5 x (0^2 +2)
15 = 5 x 3 = 5 x (1^2 + 2)
30 = 5 x 6 = 5 x (2^2 + 2)
In general, 5 x (the "middle" square plus 2);

Algebraically, this leads to a discussion of the benefits of representing 5 consecutive integers as n-2, n-1, n, n+1, n+2.

From here we could ask them to explain why or show that 2010 CANNOT be expressed as the sum of THREE consecutive squares!

Finally, is FIVE the maximum number? That is, can 2010 be expressed as the sum of N consecutive squares for some N greater than 5?

Thank you, David, for getting my brain working...

Dave Marain said...

I neglected to thank you for providing a link to the Online Encyclopedia of Integer Sequences. Some may not be aware of this resource/tool. By placing 2010 in the search box, the site finds 198 properties of 2010! Try it!

Of course, it sorta takes the fun out of the activity so I wouldn't let students in on it right away!

Eric Jablow said...


Perhaps you should ask your students this:

"Can 2010 be the sum of an even number of consecutive squares?"

After that, you can start with general formulas for the sum of consecutive squares, and you can introduce them to the identity for the sum of the first k squares, perhaps proving it by induction. We've already discussed the identity for the sum of the first k cubes.

In fact, you can find many sources of advanced problems in the modern classic, Concrete Mathematics, by Graham, Knuth, and Patashnik. You can explain to your advanced students the resemblance between the Δ operator and the derivative operator, or talk about Catalan numbers and other special number sequences. For middling students, you can discuss the Josephus problem, which will introduce them to give them a new appreciation of logical thinking.

mathmom said...

I got this off of a Mathletics Facebook status message: 2010 = 1+2-(3-4-5)*6*7*8-9

Could certainly make a good challenge problem (insert the appropriate operations and parentheses to make this a true statement...)

Anonymous said...

2010 = 2 * 3 * 5 (7 + 11 + 13 + 17 + 19)

Not my discovery though, link beneath :)

Alexey Izvalov said...

Hi! I also searched for interesting math facts about 2010. I missed two (the most interesting, as I can judge )facts, found by David. But I found that:
There are 8 ways to get 2010 as a difference of triangle numbers (T_n=n(n+1)/2)

22 ways to get 2010 as s sum of three triangle numbers

8 ways to get 2010 as a sum of 3 squares

3 ways to get 2010 as a sum of five cubes (only one way if we require the cubes to be completely different)

84 ways to get 2010 as a sum of two primes

Fraction 1/2010 can be predented as a sum of two fractions with nominator 1 by 41 ways, and a a difference - by 40 ways

Dave Marain said...

Look what this innocent problem has generated! Thanks, Eric, mathmom, sostotigog, Alexey, David and Mike for your contributions. Do you think students would invest this much effort?

Actually, I believe students would quickly search the web and some are even more adept at this than us, certainly more than me!

Some of these observations reflect human obsession with numbers. I'm currently reading "The Curious Incident of the Dog in the Night-time." The references to math, primes in particular, are fascinating. The author takes us inside the mind of the pure mathematical mind, unfettered by emotion.

mathmom said...

I posed this to my older middle schoolers, and posted some results on my blog (more to come -- some of them didn't have time to work on it and are handing it in next week). I don't think they searched the Internet based on what they came up with (though I'm guessing that one went to the internet for a list of primes).

Dave Marain said...

Thanks for sharing the experience. Here's another variation to try with them. They can do it on their own time after a brief introduction (modify it to make it more interesting!):

Goldbach's Conjecture:
Every even number greater than 4 can be written as a sum of 2 odd primes in at least one way.

For example, 2010 = 7 + 2003.
Using a table of primes, list all possible ways for 2010 to be represented.

Here's a good reference:
Goldbach Generator

mathmom said...

The Goldbach one actually sounds pretty mechanical and uninteresting to me. Take a prime, subtract it from 2010, see if the difference is prime, lather, rinse, repeat. With younger kids who needed the subtraction practice, that could be a good way to make it easier. But I don't see a good puzzle or pattern to be found in that (at least not without trying it for other numbers as well). I don't get this group back until the end of the month, at which point I think we'll do some targeted MathCounts practice.

Thanks as always for your interesting posts!

mathmom said...

Oops, I didn't notice until I re-read your suggestion that you did say to modify it to make it more interesting. ;-)

Eric Jablow said...

Goldbach's conjecture leads to some interesting mathematical discussions:

1. Vinogradov's theorem: Every sufficiently large odd number is the sum of three primes. Mathworld gives the current minimum as about 3.33×10⁴³⁰⁰⁰.

2. Chen's theorem: Every sufficiently large even number is either the sum of two primes or the sum of a prime and the product of two primes.

3. Estemann's theorem: Almost all even numbers are the sum of two primes. [The even numbers that aren't have density 0.]

Why are results like Vinogradov's fairly common?

Why do we see 'X is true for all sufficiently large integers' for many different properties?

How do mathematicians discover results with ridiculously large thresholds? Consider Skewes' numbers.

Now, none of these questions are suitable for exams, but they are suitable for discussions of mathematics.

Alexey Izvalov said...

Last year I gave my students a non-obligatory task to get 2009 as a sum and as a difference of two squares. Many of them solved it. Now I'm looking forward to meet them in the new semester and see, if they find that 2010 is the sum of 5 consequtive swquares

Unknown said...

Nice thread!

Here's another fact:
2010 is the second smallest positive integer which, read as a base-n number, is a multiple of 3, for all n bigger than 2; the smallest example is 1020. Justifying this nicely ties together several bit of high school math:
(a) different number bases;
(b) what numbers in base 10 are multiples of 3?
(c) if n is a multiple of 3, the multiples of 3 are those whose last digit is a multiple of 3;
(d) remainders mod 3 of powers of n.

With all this in hand, the proof is fairly easy. Since the number makes sense in base 3, the only allowable digits are 0,1,2. By (c), we see that the number must end in a zero. The base-10 rule then rules out all numbers below 2010 except 1110, 120, 1200, 210, 1020. Base 8 rules out the first 4 of these; checking this is easier using this base converter.

Thinking about (d) allows us to see that 1020 and 2010 are indeed examples. With a little extra effort, it is not hard to characterize all numbers with this property.

Lastly, if you want a number to be a multiple of 3 in all bases including 2, the smallest example is 101010.

Sorry for being long-winded!