Can you find all the locheerleading poster ideasops? Can you predict which loop a number might end up in?

doing there?! It seems like magic that these terms are all connected. If you look at the proof, it makes more sense. Nevertheless, there&8217;s a way that I continue to carry a sense of wonder that all these pieces fit together,againandagain, unexpectedly and inevitably.

Happy post-Thanksgiving! To all Seattlites looking for a math-rich event for kids, we&8217;ve finally got someMath Salonson the calendar, starting in January. Interested in joining us? Justrsvp, and we&8217;ll see you there.

So that&8217;s it!

The Julia Robinson Festival is challenging but noncompetitive, highlighting rich mathematical problems curated by adults who love math. Its a perfect place for students who love math, games, or puzzles; for contest-goers looking to see what else math is about; or students who prefer the collaborative, noncompetitive aspects of math to contests.

[Erdős] spoke of &8220;The Book&8221;, an imaginary book in which God had written down the best and most elegant proofs for mathematical theorems. Lecturing in 1985 he said, &8220;You don&8217;t have to believe in God, but you should believe inThe Book.&8221;

In any case, let&8217;s look at this puzzle in mod 9. This is kind of like looking at the shadow of the numbers as they travel through the dr square puzzle. They sometimes lurch up or down, but they might seem to stay the same to us if we only care about their mod 9 shadow. Let&8217;s see what happens.

Math Salons&8211;our free, community mathematical game events&8211;are back! We&8217;ll be hosting the next one at the Capitol Hill library from 4-6pm on Monday, January 23.RSVP here. Elementary teachers are especially welcomed!

We&8217;ll see you at the festival!

Hi Everyone! We&8217;ve been away for the holidays, but back now, and looking forward to diving back into the deep end. While we&8217;re gearing up, I thought I&8217;d share aquick link to a lovely puzzle without words of James Tanton&8217;s. I saw him present this at aJulia Robinson Festivalin Houston, and was impressed enough to use the puzzle frequently since then in my own teaching. The beauty of the graphics makes it especially appealing.

Step 4: Repeat steps 2 and 3 until you understand whats going on.

A thought experiment: if you have a 50 digit number, what&8217;s the largest it could be after you square it and then take it&8217;s digit sum? In the worst case, all fifty digits would be 9s, which means it&8217;s one less than a one followed by fifty zeroes (or 10^50). Squaring that latter number is easy: it&8217;s just a one followed by 100 zeroes (10^100). So our number must be less than that, which means it has at most 100 digits (and in general, squaring a number roughly doubles the number of digits it contains).

Enjoy!

But the same holds true for the bottom line and the middle line as well! So all the blue lines below must be parallel.

This is the kind of result that seems both random and astonishing. You have to draw a few quadrilaterals just to convince yourself that it even seems to hold. How do you go about proving it in general?

There are a lot of great websites on math education, but some really stand out. I&8217;m happy to report a new one which promises to be an incredible resource: mathpickle.com. In particular, Dr. Gord&8211;who runs the site&8211;has produced some beautiful activities for math classrooms that featuresolvable forays into unsolved problems. The emphasis on elegant graphical design and solid mathematics makes everything here a pleasure. This site will now have a place on our blogroll.

I&8217;m always interested in people who can successfully talk about math to a broad audience. Here&8217;s a surprise then: comedian Dave Gorman actually explains perfect friendly, and sociable numbers in this 4 minute clip from his comedy routine (note to parents: there&8217;s some language here that might not be appropriate for kids at the end).

? If you need to adjust one of them, what is the adjustment? The answer: the adjustment ctor is roughly

Notice that we have only three loops in the mod 9 world. We might call them the 0 loop, the 1 loop, and the 4 &8211; 7 loop. Might these correspond to the loops we saw before? In ct, they do: we even called them the same thing, except the 4-7 loop, which we called the 13-16 loop (notice the digit sum of 13 is 4, and the digit sum of 16 is 7). Since we know from our earlier argument that there are only three loops, we can tell from here which loop we end up in just by taking the digit sum. In other words, 187 has digit sum 16, which has digit sum 7, which means it has to end up in a loop that includes 7 (mod 9), and the only option is the 13-16 loop. So we shouldn&8217;t be surprised to see that happens when we check it out:

We&8217;re offeringcircles and classesall over town for kids grade K &8211; 10, including a class for K-1 children and their parents at the Robinson Center (Nurturing the Math Instinct), and a pair of classes for 6th and 7th graders at Seattle Central Community College on Saturday mornings (Outmaneuver!andBilliard Balls and Laser Beams&8211;registration open now!). Clickherefor the full list.

So it&8217;s a parallelogram!

James Tanton and his students just broke the folding record! Follow the link for the story!

2&8211;> 4 &8211;>

1 &8211;> 1 &8211;> etc.

For those who are interested, I&8217;ll try to prove this in slightly more detail.

There&8217;s another way to appreciate the wonder of it all, and that&8217;s with the results that connect things that seem like they shouldn&8217;t even be remotely connected. ConsiderStirling&8217;s Approximation. We could ask, how different are

Looks like it will still hold. I&8217;ll leave that one to you.

These are the kind of details that at once make math rich and, sadly, inaccessible. For me, I read the book blown away, exclaiming aloud when a new idea is dropped into the mix, all the time with that sense of &8220;how did anyone think of this?&8221; I suppose

Step 1: Choose a starting number.

Today, the solution to the puzzle.

Some students asked me why this was true the other day. I had totally forgotten how to approach the problem, so I got the chance to play around with it fresh. I had two ideas of how to start. The first was to draw another line in the drawing and see if that helped.

Now here&8217;s the thing: this is a book that is decidedlynotfor the layperson. To read it, you need to be extremely comfortable with calculus and limits, infinite sums, and some serious subtleties, both conceptual and notational. But if you are, the surprises keep coming. The results on primes are nothing short of astonishing (especially to me, as a non-number theorist who&8217;s always loved the subject).

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It sure looks like a parallelogram, doesn&8217;t it? The amazing ct here is thatno matter what quadrilateral you start with, you always get a parallelogram when you connect the midpoints.

What we&8217;ve just proven is that the line you get by connecting the midpoints is the same as the line you get when you draw a parallel through one of the midpoints. Armed with this observation, we&8217;re ready to prove our main result (which we did above).

Step 2: Square the number, and then add 1.

As always, it&8217;s a pleasure to be a part of the mathematical lives of you and your children. Hope to see you in January!

that&8217;s what happens whenever you&8217;re looking at great art. How did Stravinsky compose The Rite of Spring? How did Picasso make Guernica? It takes a huge amount of work to get a sense of how they thought of it (though there&8217;s always a path). It&8217;s inspiring, and also leads to the despair that comes from idealism. There&8217;s a prime between n and 2n! Wow! And no one knows if there&8217;s always a prime between

Step 2: Square the number.

Now this is interesting. First of all, it&8217;s much r&8211;which makes sense, since we&8217;ve cut out a lot of information. Second of all, we seem to get repeated numbers. But that makes perfect sense. We already knew that taking the digit sum doesn&8217;t change a number&8217;s mod 9 shadow, so we&8217;ll have to get repetition. This means that the only chance to change the mod 9 shadow comes when we square the number. What are the options when we square the numbers mod 9? Well, there&8217;s not much work to do to find the loops. Of course, we have to remember to cross out the numbers above 9 and replace them with the mod 9 shadows.

5 &8211;> 7 &8211;> 7 &8211;> 4 &8211;> 4 &8211;> 7 &8211;> 7 &8211;> &8230;

&8220;Beauty is the first test: there is no permanent place in the world for ugly mathematics.&8221; &8211;G. H. Hardy

Proof. I&8217;ll leave the details to you, but briefly, parallel lines produce like angles when they intersect the same line, so we can get that the top is similar to the entire , because their angles are the same size.

Addendum: Proof that the blue lines are parallel.

187 &8211;> 34969 &8211;> 31 &8211;> 961 &8211;> 16 &8211;> 256 &8211;> 13 &8211;> &8230;

Recall the puzzle:

Step 3: Sum up the digits of that number.

Step 3: Sum up the digits of that number.

Reserve your spot at the festival.Click here.

and recall that we had three discovered loops, which we called the 1 loop, the 9 loop, and the 13-16 loop. The question was: are these the only loops?

What: The Julia Robinson Math Festival, a noncompetitive event featuring an abundance ofinspiring mathematical challengesand puzzles that require creativity and persistence to solve.

Here&8217;s why: digit sums have a very nifty quality: they tell you when a number is divisible by 9 (remember the divisibility rule?). For example, 5904 has a digit sum of 18, which is divisible by 9&8230; therefore 5904 is divisible by 9. Nifty, right? (Now any curious seeker of mathematical patterns will immediately want to know how this could possibly be true. I&8217;m not going to answer that question here, but think about it. You can also checkthisout.)

Now talking about remainders when you divide by 9 is so annoymath for love cheerleading poster ideasing to type that I&8217;d like to shorten it. Let&8217;s call a numbers remainder when divided by 9 that number &8220;mod 9.&8221; So 19 = 1 (mod 9). Also, 496 = 1 (mod 9). (I&8217;d like to say 496 = 19 (mod 9) as well, since that&8217;s how I like equals signs to work&8230; but I suppose that my definition doesn&8217;t quite allow it. Or does it? Is there a better definition I could have used?)

. The proof is too involved to get into here, but consider the depth of that statement: on one side, you&8217;re going up number by number, multiplying primes when you come to them, ignoring nonprimes; on the other side, you&8217;re putting in a ctor of 4 for each number. No matter how large the primes get, the product of fours is always bigger.

Proofs from THE BOOKis a collection to these &8220;book proofs,&8221; some of the most elegant arguments in mathematics. One thing I like about this book is that it samples multiple, very diverse ideas to prove the same, often classic, result. For example, the book opens with six different proofs that there are infinitely many prime numbers. A number of these are classics, or variations on a classic: assume that there are only finitely many, and then find a contradiction by constructing a new prime, or showing one must exist. But some of the constructions are incredibly novel.

5 &8211;>

0 &8211;> 0 &8211;> etc.

Step 1: Choose a starting number.

In other news, Randall Munroe recently created thisincredibly dense poster of information, tracking money, how much of it there is, and what it gets spent on. It&8217;s quite a feat, and worth looking at in detail.

I think the DR Square Plus One is an even neater puzzle than the previous. The solution is even a bit tidier than it&8217;s predecessor.

? I&8217;ll solve that! And then, of course, you realize just how hard these unsolved problems are. It&8217;s easy to go from surveying the best work to deciding that you don&8217;t have what it takes. Still, these works of art are worth seeing.

We&8217;re nowpartnering with schoolsto offer teachers professional development, curriculum enhancement, and other mathematical support. It&8217;s been a privilege and a pleasure. If you&8217;re interested in bringing us in to your school, let us know.

math for love cheerleading poster ideas,Who:4th-12th graderswho are interested in math, games, and puzzles; students who prefer thecollaborative, noncompetitive aspects of math, or contest-goers looking to see what else math is about; parents and teachers are also welcome.

There are a limited number of spots.Register now.

It&8217;s the same as the picture from above! Can you see it? Let&8217;s erase the bottom half of the picture, and make the lines that are parallel the same

Now let&8217;s take the digit sum. The greatest digit sum a hundred digit number could have is 9+9+&8230;+9 one hundred times, or 900. Meaning if we start with a 50 digit number, we&8217;ll be down to a three digit number in no time. From three digit number, we square to six digit number, and then take the digit sum to at most 54. If you follow this logic, you can see that no matter how large a number you start with, you&8217;ll end up below 60 (or below 30 or so, if you want to continue) at some point, so all you need to do is check the first 60 (or 30 or so) numbers and see if they end up in one of those loops. I&8217;ll leave the details to you, but I&8217;ve checked, and there are indeed only three loops that numbers less than 60 end up in, so hence only three possible loops.

First thing: what we don&8217;t want to do here is check every number; that would take, literally, till the end of time. So our first priority in this case is to limit the amount of work we need to do. There&8217;s a nice way to do this, it turns out, and it hinges on noticing that for most large numbers, taking a digit sum reduces the size of a number much ster than squaring a number increases its size.

1 &8211;> 1 &8211;> 1

Lemma. Take a , draw in the midpoint P as show above, and draw in a line segment from P parallel to the base of the . The point Q, where the line segment meets the , is also the midpoint of its side.

I&8217;ve been readingProofs from THE BOOK by Aigner and Ziegler, a paean to beautiful mathematics.The Bookrefers to a conceit of the mathematicianPaul Erdős. From wikipedia:

Here&8217;s a followup puzzle, which I call theDR Square Plus One.

For those who are interested, here&8217;s the higher power mathematical method to show that you don&8217;t need to test numbers higher than 30 (in ct, 26) to find the loops in the DR Square problem. Let x be your starting number. Your next two steps will take you to x^2, and then to something less than 9log(x^2) = 9(2log x) = 18 log(x). (Here log is log base 10.) We can see that 18 log(x) > x will be true only if x is sufficiently small, and playing around with this will give us a bound. For example, x = 100, gives 18log(x) = 36, which means we can assume that x must be less than 36. Iterating or graphing gives us the final result that we only need to check up to x < 26.

In ct, digit sums are even better: they don&8217;t change the remainder of a number divided by 9. For example, 5904 had remainder 0; so did its digit sum. 496 has the same remainder as its digit sum, which is 19, which has remainder 1 when divided by 9.

0 &8211;> 0 &8211;> etc.

For example, there are these incredible inequalities throughout the book. Bertrand&8217;s Postulate states that there is always a prime number between any number n and its double 2n. I&8217;ve always thought that this was a great result. But the proof is magical: a series of inequalities that leads to a contradiction that a quantity ends up growing larger than it should, unless there&8217;s a prime between n and 2n. On the way, there&8217;s this elegant proof of the remarkable ct that the product of all primes less than a number x is less than or equal to

(and in the shadow version mod 9, where 41 = 5)

I found this quite a pretty line of argument: drawing in the lines from opposite corners turns the unthomable into the (hopefully) obvious. That resolution from confusion to clarity is, for me, one of the greatest joys of doing math.

Know others who would be interested in joining us?Please pass the word along to them.

(Click here forPart IorPart II)

At the festival, students will have an entire morning to explore a number of scinating problems in mathematics with the help and encouragement of our volunteers. There will be tables, staffed by one or two volunteers, set up to introduce the students to a mathematical problem or puzzle that requires creativity and exploration to solve. The activities are designed to engage students for about 20-30 minutes. These activities will range from levels that older elementary students will enjoy, to levels that will challenge the brightest high school students.

But the left side of the top is half of the corresponding side of the entire . That means the same must be true of the right side. In other words, we know that Q must be the midpoint of its side. Done!

It sure looks like connecting those midpoints creates four congruent s, doesn&8217;t it? In ct, that&8217;s not too hard to prove. Once we know that, we can see that any pair of touching s forms a parallelogram. That means that we have the two blue lines below are parallel.

Step 4: Repeat steps 2 and 3 until you understand whats going on.

41 &8211;> 1681 &8211;> 16 &8211;> 256 &8211;> 13 &8211;> 169 &8211;> 16 &8211;> (13-16 loop)

That&8217;s a pretty nice argument, I think. There&8217;s more to this problem, though, like, is there any quick way to tell which loop a number ends up in? The answer is yes: it turns out that if you just take the digit sum over and over and forget squaring, it won&8217;t affect which loop you end up in. In other words, 7561 ends up in the same loop as its digit sum, 19, which ends up in the same digit sum as its digit sum, 10, which ends up in the same loop as its digit sum 1. In other words, 7561 ends up in the 1 loop.

Here&8217;s what it looks like for an arbitrary .

4 &8211;> 7 &8211;> 4 &8211;> 7&8211;> etc. (we saw this loop already, up two lines)

Doesn&8217;t it look like the blue line is parallel to the red lines above and below it? If that were true, that would give us a powerful way forward. It also presages my second idea: try connecting the midpoints of a rather than a quadrilateral.

The next question is whether we can break the result by pushing back on the initial setup. Does our result hold, for example, when the quadrilateral isn&8217;t convex?

See that the blue lines are parallel? The top line connects the midpoints of a , so we can apply our lemma!