## Another TI-84 Plus Silver Edition giveaway puzzle contest!

I've been thoroughly enjoying the books James Tanton sent me, which infuse Math education with life and joy and play, along with the Youtube videos that cover some of the topics in the books. One of his videos particularly intrigued me because it covers a topic I've never run into in my 30+ years of playing with Math. So, I thought I'd invite each of you to watch this particular video, to solve the challenge at the end of the video and, if you're lucky, to win a TI-84 Plus Silver Edition calculator for your efforts.

But first, a word from our sponsor:

STEM skills - science, technology engineering and math - hold the key to tomorrow's innovation. To help students learn tomorrow's job skills, Wild About Math! has teamed-up with Texas Instruments to give away a TI-84 Plus Silver Edition graphing calculator with GraphiTI package. To learn more about TI-84 graphing calculators or other TI products, visit: http://education.ti.com. To learn more about STEM careers, visit: http://education.ti.com/studentzone.

## TI-84 calculator giveaway winner

[ Sorry for the delay in announcing the winner. I ran into a technical issue that I needed to get resolved.]

On November 1st I announced a contest. Folks entering the contest were eligible to win if they correctly solved a Math problem of my choosing. The prize was a TI-84 calculator, graciously donated by Texas Instruments.

Random.org selected Nate Burchell as our winner. We had 20 submissions. Almost all of them were correct.

I sometimes post a solution other than the winner's because sometimes the winner's solution, while correct and well explained, is not the best example of how to solve a problem. In this case, though, Nate explained his solution quite well.

Here is Nate's solution.

Nate is a Math teacher and blogs at http://burchellmath.blogspot.com.

Thanks to everyone who played. Stay tuned for another puzzle or three in the coming weeks.

## TI-84 Plus Silver Edition giveaway puzzle contest

**The contest is over. I've announced the winner here.**

[ Update 11/2. I changed the title of the post to say "TI-84 Plus Silver Edition" rather than "TI-Nspire." ]

No, I'm not reviving Monday Math Madness on a regular basis but Texas Instruments (TI) has offered calculators and t-shirts to readers so I'm going to run a few puzzle contests and give away a prize to the winner of each contest.

Before I announce the contest and rules, here's a word from our sponsor:

STEM skills - science, technology engineering and math - hold the key to tomorrow's innovation. To help students learn tomorrow's job skills, Wild About Math! has teamed-up with Texas Instruments to give away a TI-84 Plus Silver Edition graphing calculator with GraphiTI package. To learn more about TI-84 graphing calculators or other TI products, visit: http://education.ti.com. To learn more about STEM careers, visit: http://education.ti.com/studentzone.

## MMM #38: We have a winner!

Random.org selected Ken Lareau as the winner for MMM #38. Congratulations, Ken! Here's what the problem was:

Prove or disprove: The product of any four consecutive integers is always one less than a perfect square.

Don’t assume the integers are all positive.

Here's Ken's solution:

The above conjecture is true. The domain can be simplified a bit by noting two things:

1) If there are both positive and negative integers in the product, then since they're consecutive, 0 has to be present, meaning the product is 0, which is one less than 1, a perfect square.

2) If all the numbers are negative, their product is positive and equal to the product of the negation of each number (in other words, (-a) * (-b) * (-c) * (-d) = a * b * c * d).

Therefore we can restrict our validation to consecutive positive integers. From this, the proof is easy to see since if the first number is 'n', then we have for the product:

n * (n + 1) * (n + 2) * (n + 3)

which expanded is:

n^4 + 6n^3 + 11n^2 + 6n

One more than this is:

n^4 + 6n^3 + 11n^2 + 6n + 1

which happens to factor down to:

(n^2 + 3n + 1)^2

which is a perfect square. Hence the conjecture holds for all possible products.

I have one comment on the problems and on the solutions submitted. Many of the submissions ignored the part about not assuming all the integers to be positive. While it may be trivial to many of you that the signs of the numbers doesn't affect the solution, if you ignore that part of the problem then you're not solving the whole problem. Ken did a superb job of showing the signs don't matter. Yes, I realize that the algebraic solutions don't assume anything about the signs of the numbers but it's important to make that point in your argument.

And, here's a solution at Watchmath.com

## MMM #37: Prizes 2 and 3

Today I'm announcing two more prizes for MMM #37.

Here is IB Yea's solution:

The problem is to find the algorithm for the coordinate and number for the spiral that goes like the following:

25 24 23 22 21

10 09 08 07 20

11 02 01 06 19

12 03 04 05 18

13 14 15 16 17Basically, this is how the spiral goes:

From the starting point, it goes -1 via x, then -1 via y, then 2 via x, 2 via y, then -3 via x

and it repeats itself like that so that:x-0 + -1 + 2 + -3 +...

y-0 + -1 + 2 + -3 +...This comes from the fact that the sqare enveloping the square below always has its number starting

below the top left number, the highest one, which follows (2n+1)^2.

If so, then it has to go 2n-1 down (since the number below the top left one is included), 2n to the right (enveloping the bottom part of the square below and jutting out),

2n up, then 2n+1 to the left to completely cover the square below and start the next square. Also, the one previous to 2n-1 down is 2n-1 left because of the smaller square.So basically,add up the numbers from the sum above to find x and y.

The sum above is the alternating series:

Sum n-(1 to n) ((-1)^n)*nThe equation above can be used in the following condition:

(1)The equation above for x and y, n can be the same,

or (2) n in x equation is 1 higher than n in y.

With those conditions, you can proceed to find the numbers in between by just adding numbers in x for condition (1)

or by adding numbers in y in condition (2)Then, using those same numbers from the sum, plug it in the equation below and add it in the following way:

n=1+(sum |x|)+(sum |y|)

For example:

x=-1+1=0 n=1+(|-1|+|1|)+(|-1|)=3

y=-1x=5=-1+2-3+4-5+6-7+8-9+10

y=2=-1+2-3+4-5+6-7+8-9+7

Plugged it in n=1+(|-1|+|2|+|-3|+...+|10|)+(|-1|+|2|+|-3|+...+|7|)=1+55+52=108

In the above case, a summation formula helps.Albeit this method is really bad if you want to use large numbers. 🙁

Although it helps if you use 2a=n (in this case, n is the term in the series, a is the "answer")

since it alternates like this: -1, 1, -2, 2, etc

Watchmath has solutions posted here and here.

Stay tuned for a few more solutions and prizes.

## MMM #37: Prize #1

I'm going to give a number of $10 prizes for MMM #37. Here's the first to Alec Cooper for his very elegant solution. I paid Alec to look through all the submissions to determine which were correct. Most were. I'll be giving a prize to everyone who turned in a correct solution and I'll be publishing each of them as well, one or two per posting. No, I won't be giving double prizes to folks who solved both problems.

Alec solved both problems and even sent me some Java code that performs the second algorithm.

Here are Alec's solutions.

Here's part 1:

And, here's part 2:

Stay tuned for more postings of solutions and more prizes.

## MMM #38: square or no square?

Results of MMM #37 should be out in the next couple of days.

Here's MMM #38:

Prove or disprove: The product of any four consecutive integers is always one less than a perfect square.

Don't assume the integers are all positive.

## Still working on reviewing MMM #37

I got submissions from eight people for this latest problem. Yes, this problem was tougher than most so I'm delighted to have gotten so many entries. I recently started work on a big software development project on a very tight schedule so I've not had the time to check each solution for correctness and that's what's delaying the announcement of winners. I'm paying one of the submitters to check all of the entries. What he'll probably do is run the algorithms against a set of numbers and confirm that they produce the expected answers. In a few days I should be ready to award some prizes.

In the meantime, you are welcome to post your solution on your personal blog if you have one. I'm always happy to link to people's blog entries but please don't ever post your solution until after submissions are due. For this particular contest I will publish all correct solutions.

Even though I'm behind in reviewing this contest I'll still post MMM #38 tomorrow. Sorry for the delay.

## MMM #37: More spiral fun

Our new Monday Math Madness extends the exploration of MMM #36.

Here's the problem:

Based on the introduction to spiral numbers presented in MMM #36, solve one (or both) of these problems:

- Come up with an algorithm that tells what number is at an arbitrary X, Y coordinate.
- Come up with an algorithm that tells the X, Y coordinates for an arbitrary positive integer.

I'll give three prizes for this contest, one to a random solver as usual, and one each for the best solution to each part of the problem. Even if you've won a prize in the last year you're eligible for one of the "non-random" prizes.

## MMM #36: Spiral Numbers – Winner!

27 of you submitted solutions to MMM #36. Random.org has selected Mathias Malandain as the winner. Congratulations, Mathias!

This was the problem:

Imagine arranging the positive integers in a spiral pattern.

The numbers from 1 to 16 look like this in the spiral pattern.10 9 8 7 11 2 1 6 12 3 4 5 13 14 15 16The location of each number corresponds to an X,Y Cartesian coordinate where the number 1 is at the origin: (0,0).

2 is at (-1,0). 3 is at (-1,-1). 4 is at (0,-1). 5 is at (1,-1). 6 is at (1,0). 7 is at (1,1) and so on.What is the X,Y coordinate of the number 1,000,000?

Show your work.

Most people solved the problem by noticing where squares lie in the spiral.