To all those studying trigonometry — let’s take the mystery out of special right triangles, shall we?  In a trig class, students are expected to know the sine and cosine of angles that are 45 degrees, or perhaps a multiple of 45 degrees.  Here is my effort to make it as simple as possible to “see” what these values should be, without having to resort to memorization. We start with a right triangle with a 45 degree angle.  For convenience, let the hypotenuse have length 1.  We are dying to find out how long the legs are.  To find out, we apply the Pythagorean Theorem, and we get this: Next we go ahead and square things, giving this: Time to get clever: since one of the acute angles is 45 degrees, the other one must be as well (how else can you get 180 degrees?).  And a triangle with two equal angles must have two equal sides, so — guess what? — the legs of the triangle are equal.  And the value of “?” that satisfies ? + ? = 1 is clearly 1/2. Therefore, in a right triangle with a 45 degree angle, if the hypotenuse is 1, then the legs are both the square root of 1/2.  That’s it!  Once you’ve done these mental steps a few times, it should be fairly easy to close your eyes and come up with the solution in a matter of seconds — and there is really no need to memorize anything!  Just apply the Pythagorean Theorem, and it’s very quick to see that you need to get 1/2 + 1/2 = 1 once you’re done squaring things.

Next post: analyzing 30-60-90 triangles, with no memorization!

How do you teach students to multiply numbers when one or both of them is negative?  This is a surprisingly challenging question.  In this post I offer some thoughts about how to accomplish this in a way that is mathematically legitimate and yet still easy to grasp.

Notion #1: What is a negative number?

Just what is a “negative number,” anyway?  Just to be quirky I’ll refer to “anti-numbers,” such as 3 and anti-3.  (Thinking of matter and anti-matter.)  The diagram below shows 3 piles of sand and 3 anti-piles, or more simply, 3 holes. The important thing to know here is that we can squish a pile into a hole, and when we are done, we are left with the same flat sandbox we started with before building any piles or digging any holes.  So here we have a visual representation of the idea that 3 and anti-3 “cancel out.”  In symbols, we write 3 + -3 = 0.

Notion #2: Multiplication as Repeated Addition

When students first encounter multiplication, they learn to compute 5×3 by writing out 5 groups of 3, which gives 15: Problem #1: What is 5 x -3?

Using Notion #2, we think of this as being 5 groups of anti-3.  The diagram below shows that this is simply anti-15: Here we are assuming that students know already how to add negative numbers.  To read about how to develop this notion using piles and holes, see here.

Okay, so (positive) x (negative) turned out to be a piece of cake!

Problem #2: What is -5 x 3?

When we try to interpret this using Notion #2, as we did with Problem #1, things get a bit awkward: what is “anti-5 groups” of 3?  It’s not at all clear what this should mean, so we are going to have to appeal to a new technology called…

Notion #3: Multiplication Distributes over Addition

Michael drew 5 groups of 4 dots on a piece of paper, then tore it into 2 pieces as shown below: On piece A, there are 3 groups of 4 dots.  On piece B, there are 2 groups of 4 dots.  Together, there must be 5 groups of 4 dots.  The diagram below shows how to represent all of this mathematically: As a check, we can see that 12 dots and 8 dots make 20 dots altogether.

Problem #2 Revisited!

Let’s try to tackle Problem #2 with the help of high-tech Notion #3.  What exactly is anti-5 groups of something?  Well, perhaps it is, in some sense, the opposite of 5 groups of something.  Let’s see where this idea takes us: I want to know what anti-5 groups of 3 is.  I already know that 5 groups of 3 make 15.  If the distributive property, which holds for positive numbers, is also to be believed for negative numbers, then we can work out the answer to Problem #2: 5 groups plus anti-5 groups must make no groups at all, as the diagram above shows.  But if we also believe that “zero groups of something” makes zero, then the answer is staring at us!  If 15 + ? = 0, then the missing quantity must be anti-15.  Thus we would like to believe that -5 x 3 = -15.  That is, if we believe the distributive property still holds for negative numbers, we are forced to conclude that -5 x 3 = -15.  Problem solved!

Problem #3: What is -5 x -3?

Here we have anti-5 groups of anti-3.  Wow!  This one looks like a toughie.  Invoking Notion #3 again, we get this puzzle: It looks as though the distributive property is telling us that anti-5 groups of anti-3 is simply the opposite of 5 groups of anti-3, which we know from Problem #2 is anti-15.  We are again faced with zero groups of something, and we are committed to the belief that 0 x -3 = 0, so the solution to -15 + ? = 0 must be anti-anti-15!  Whoa — things are getting a bit crazy — what exactly is anti-anti-15?  Well, since 15 + -15 = 0, it must be the case that anti-anti-15 is just 15!  So once again our prior belief in the distributive property necessitates the conclusion that -5 x -3 = 15.

Summary

1. To compute 5×3, we use the notion of “repeated addition” to get 5×3 = 15.  This means that (pos)(pos) = pos.
2. To compute 5x-3, we can again use repeated addition, giving 5x-3 = -15.  This shows us that (pos)(neg) = neg.
3. To compute -5×3, we invoke the distributive rule to see that the answer must be -(5×3) = -15, so that (neg)(pos) = neg.
4. To compute -5x-3, we also use the distributive rule.  This shows us that the answer must be -(5x-3) = -(-15) = 15.  So (neg)(neg) = pos.

That’s it!  All the sign rules for multiplication can be worked out using straightforward logic, provided we understand 1) what a negative number (or anti-number) is, 2) how to perform addition with both positive and negative numbers, and 3) how the distributive rule works.  (Technically, we also used the fact that (0)(a) = 0 and the fact that additive inverses are unique, but surely these are minor details.)

Here is a standard textbook question: find an exponential function y = a*b^x whose graph passes through the points (1,6) and (3,24).  The standard presentation produces a formula that is indeed correct, but the process used to arrive at it is not very intuitive, and does not give students a great deal of insight into the nature of exponential functions.  But there is a better way!  I’ll develop some ideas below, and I hope you’ll agree.

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I love to use this 15-second video clip when exploring exponential growth with students:

Can we write a formula to describe the growth of the bacteria in the video?  Assuming we start with just one bacterium, how many bacteria will there be in, say, 5 hours?  And how do we model other, similar situations using exponential functions?  Let’s explore.

I’ll stick with bacteria growth for the sake of giving students something concrete to keep in their minds as we do the math, but let’s allow the growth rates to be whatever we want (not necessarily “doubling” all the time).

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Scenario #1

At 12:00 PM a scientist observes 20 million bacteria in a petri dish.  At 1:00 PM, she observes 60 million bacteria.  How many bacteria can she expect to see by, say, 6:00 PM?

Solution: The boring version of this question is this: find an exp. function that contains the points (0,20) and (1,60).  Here we start timing at 12:00 PM, so t = 0 at that time.  We can figure out without a great deal of fuss that the answer must be y(t) = 20*3^t.  Students should check using mental math that the two given points satisfy this equation.  Two important observations: we had initially 20 million bacteria, and the number 20 is proudly displayed in the equation.  Also, we can see that the number of bacteria tripled in one hour, and again the number 3 is on display in this equation.  All well and good so far, but let’s play with this scenario a bit!

Scenario #2

A scientist observes 20 million bacteria at 4:00 PM.  By 5:00 PM, there are 60 million bacteria.  Find a function that models these data.

Solution: Now we want the function to include (4, 20) and (5,60).  Let’s be a little bit clever and *set our clocks back.*  If instead of using t we use t – 4, we can instantly see that this function does the trick: y(t) = 20*3^(t-4).  Again, students can use mental math to check that the two given points satisfy this equation.  And students should still observe that the number of bacteria tripled in one hour.  The equation displays the amount of bacteria at t = 4, the fact that the number triples in one hour, and the fact that we have set our clocks back 4 hours.   That wasn’t too bad!  Let’s make the problem a touch harder.

Scenario #3

A scientist observes 20 million bacteria at 12:00 PM.  By 5:00 PM, there are 60 million bacteria.  Find a function that models these data.

Solution: This time we need the function to include (0, 20) and (5, 60).  And now we can use a second clever trick: let’s *compress time!*  This time it took 5 hours to triple the number of bacteria.  But instead of thinking of this as 5 of something, why not instead think of it as 1 of something?  We can do this by organizing time into 5-hour intervals.  How is this accomplished?  Just use t/5 instead of t!  Surely the two given points satisfy the equation f(t) = 20*3^(t/5), as students can check mentally.  Once again we see from the equation the role that “20 million bacteria” play, the role that “tripling” plays, and as a bonus we see that we have compressed time by a factor of 5, treating 5 hours as 1 block of time!

Once students are comfortable setting the clock back, and are comfortable compressing time, it’s time to bring both features together at once!

Scenario #4

A scientist observes 20 million bacteria at 4:00 PM.  By 9:00 PM, there are 60 million bacteria.  Find a function that models these data.

Solution: First we observe that it took 5 hours for the number of bacteria to triple.  Now we seek the function that includes (4, 20) and (9, 60).  Clever move #1: set the clock back so that 4:00 PM behaves like noon: use t – 4.  Clever move #2: compress time so that 5 hours feel like one long block of time: use t/5.  But wait, we are not using t, we’re using t-4!  So make that (t-4)/5.  With these simple tricks, we quickly arrive at the answer: y(t) = 20*3^(t-4)/5.  It is still a matter of doing simple mental math to check that (4,20) and (9,60) satisfy this equation!  And we can clearly see the role that the number 20 million plays (we knew there were that many at 4:00 PM), the role that tripling plays (the ratio of 60 million to 20 million), the fact that we delayed time by 4 hours (since we started at 4:00 PM), and the fact that we compressed time by a factor of 5 (since 9:00 PM is 5 hours after 4:00 PM).  Brilliant!

Scenario #5

A mathematician observes that y = 19 when t = 13, and that y = 25 when t = 23.  He seeks an exponential function that fits these data, and he needs our help!

Solution: using the techniques described above, we have the answer in a jiffy!  y(t) = 19*(25/19)^(t-13)/10.  Check for yourself that this function includes (13,19) and (23,25).  This function was created to make checking these two points as easy as possible!

Observations about the equation y(t) = 19*(25/19)^(t-13)/10:

• 19 is the y-value when t = 13, the “first observation.”
• 25/19 is the ratio of the two known y-values.  This conveys something about the rate at which the y-values are growing.
• t – 13 is a convenient expression, since the first point has t = 13.
• (t – 13)/10 is also convenient, since the space between t = 13 and t = 23 is “10 hours.”

That’s it!  Once students have enough practice, they can get their answers with little or no work, and with complete understanding!

Summary:

• Set the clock back!  Use a transformation of t into t – (something).
• Compress time!  Change t into t / (something).
• Now you are ready to rock and roll!  The math becomes easy once you do those two things.  Answers pop out in about two or three lines of writing!

Thanks to Dr. James Tanton for inspiring this essay.  You can see his treatment of this subject here.

Here’s another effort at tackling instruction around counting problems.  Inspired by James Tanton’s great work over here!

You have 5 players on a basketball team.  We shall refer to them by their jersey numbers: {1,2,3,4,5}.

1. The team is travelling to an away game on a bus.  The players will sit in a line, one in front of the other.  How many seating arrangements of these 5 players are possible?
2. The coach needs to assign positions to each of these players.  2 of them will be guards and 3 of them will be forwards.  How many assignments are possible?
3. The coach needs to pick a point guard and a shooting guard.  How many assignments are possible?

There are perhaps several ways to approach these problems.  We seek methods that satisfy these criteria:

• Generalization.  We prefer a method that can be used to solve other, similar problems.
• Unification.  We prefer a method that will unify the three scenarios, so that we can treat them all just as “counting problems!”

Scenario #1: Seating Arrangements

 Seat   A Seat   B Seat   C Seat   D Seat   E 1 2 3 4 5 1 2 3 5 4 1 2 4 3 5 1 2 4 5 3 … … … … …

Analysis:

• First we asssign a player to seat A – we have 5 choices.
• Next we assign a player to seat B – since one player is assigned to seat A, we have 4 choices remaining.
• In the same way, there are 3 choices for C, 2 for D, and only 1 left for E.
• Thus there are 5*4*3*2*1 seating arrangements, or 5! = 120 possibilities.

Scenario #2: Assigning Positions

• We’ll use “F” for forwards and “G” for guards.  This is starting to feel like a labeling problem!
 FFF.GG FFF.GG FFF.GG FFF.GG FFF.GG FFF.GG Row 1 123.45 132.45 213.45 231.45 312.45 321.45 Row 2 123.54 132.54 213.54 231.54 312.54 321.54 Row 3 124.35 142.35 214.35 241.35 412.35 421.35 Row 4 124.53 142.53 214.53 241.53 412.53 421.53 Row 5 … … … … … … Row 6 … … … … … …

[Note: these tables were originally developed in Word, and the rows were color-coded.  If anyone wants the file, email me and I’ll send it to you.  So in the text that follows, the “blue-block” is the first two rows, the “yellow-block” is the next two rows, etc.]

Analysis:

• If we simply continue to list every combination of players as we did in the first problem, we would see something like the table above.
• In that case we would again have a table with 120 cells, but scenario #2 certainly has fewer outcomes than this.  What are we to do to correct this?
• Let’s examine the table very closely.  In rows 1 and 2, players {1,2,3} are forwards, and {4,5} are guards.  In row 1, the guards are in the same “seats” and the forwards are shuffled.  Since we are shuffling 3 players around, there are 3*2*1, or 3! = 6 arrangements.
• In row 2, the forwards are in the same “seats” as row 1, but the guards have been shuffled.  Since there were 2 players to shuffle, there are 2*1, or 2! = 2 arrangements.
• Summing up: in this scenario we are only interesed in the positions assigned to each player, so all of the blue arrangements are *equivalent,* all of the yellow are equivalent, and so on.
• There are (3!)(2!) cells in the blue block, and also the yellow block, and so on.
• Thus to count the number of player-assignments possible in this scenario, we take the number of cells in the whole table, which was 5!, and divide by the size of each color-block, which is 3!2!.  So we believe there ought to be 5! / (3! 2!)ways to choose 3 forwards and 2 guards from a group of 5 players.  That gives 5*4 / 2 = 5*2 = 10 choices for the coach this time around!

Scenario #3: Choosing Guards

• We’ll use “P” for point guard and “S” for shooting guard.  But what about the other players?  We’ll just use “N” for “not a guard,” but they are probably forwards!
 NNN.P.S. NNN.P.S. NNN.P.S. NNN.P.S. NNN.P.S. NNN.P.S. Row 1 123.45 132.45 213.45 231.45 312.45 321.45 Row 2 123.54 132.54 213.54 231.54 312.54 321.54 Row 3 124.35 142.35 214.35 241.35 412.35 421.35 Row 4 124.53 142.53 214.53 241.53 412.53 421.53 Row 5 … … … … … … Row 6 … … … … … …

Analysis:

• The analysis runs similar to scenario #2, but this time we make a distinction between types of guards: are you a point guard or a shooting guard?  What are the implications of this?
• First we note that the cells in the blue block are equivalent with respect to the “non-guards.”  However, this time the light blue cells in row 1 are different from the dark blue cells in row 2, since assigning player #4 to be the PG and #5 to be the SG is different from putting #5 at PG and #4 at SG.
• We conclude that the cells in a given row are “the same,” but the rows are all distinct cases.
• Thus we have 5! / 3! = 5*4 = 20 ways to accomplish task #3.
• In fact, if we consider that there are “1 arrangements” in each column for the point guard, and “1 arrangements” for the shooting guard, then we have an expression in which every player is properly labeled: 5! / (3! 1! 1! 1!).

Synthesis

Let us now revisit the original problems, and try to solve all of them with a single strategy: they are simply labeling problems!

You have 5 players on a basketball team.  We shall refer to them by their jersey numbers: {1,2,3,4,5}.

#1. The team is travelling to an away game on a bus.  The players will sit in a line, one in front of the other.  How many seating arrangements of these 5 players are possible?

Solution: There are 5! arrangements, and there are 5 labels – one for each seat: {A, B, C, D, E}.  Thus there are 5! / (1! 1! 1! 1! 1!) = 120 outcomes.

#2. The coach needs to assign positions to each of these players.  2 of them will be guards and 3 of them will be forwards.  How many assignments are possible?

Solution: There are 5! arrangements, and there are 2 labels – {G,G} and {F,F,F}.  Thus there are 5! / (2! 3!) = 10 outcomes.

#3. The coach needs to pick a point guard and a shooting guard.  How many assignments are possible?

Solution: There are 5! arrangements, and there are 3 labels – {P}, {S}, and {N,N,N}.  Thus there are 5! / (1! 1! 3!) = 20 outcomes.

James Tanton treats counting problems over here.  He argues that we never need ask “Is order important?  Is this a combination problem or a permutation problem?”  He suggests that these questions merely lead to confusion, and that drawing a distinction between the two “types” in the first place is a distraction.  Instead, he offers a unifying concept that allows us to treat all sorts of counting problems in the same manner, with clarity and simplicity.  Just ask: “How shall I label things in this problem?”

Here is my contribution to the labeling project.  For a more detailed treatment, pop over to gdaymath.com.

10 athletes are competing in the slope style event.

1. Medals will be awarded to the 3 highest scorers.  How many outcomes are possible with respect to this condition?
2. The medals are gold, silver, and bronze.  How many outcomes?
3. If the organizers decided to issue 1 gold medal, 3 silver medals, and 4 bronze medals, how many outcomes would be possible?
• First observation: if one were to write out the name-space of the athletes in every possible order, one would get a giant table with 10*9*8… = 10! entries.  We can count the outcomes for each of the three seemingly different scenarios above just by selecting appropriate groupings from this table.
• In the first scenario, 3 athletes will be called “medalists” and 7 will be called “participants.”
• Let’s take the first three athletes: Al, Bill, and Cal.  In the name-space, we would deem ABC, ACB, BAC, BCA, CAB, and CBA “equivalent with respect to condition #1.”  That is because we have simply called them “medalists.”  So the total number of entries — 10! — is too large by a factor of 6.  That is, with 3 names, there are 3*2*1 = 6 ways to organize them, all of which are “the same” in this scenario.  So our revised estimate is 10! / 3!.
• But we haven’t gone far enough: each entry in our giant table has 10 names, and we’ve only considered how to shuffle A, B, and C about.  But how many entries start ABC…?  We still have those remaining 7 athletes to deal with.  So we’ll have ABC | DEFGH(IJ), and then ABC | DEFGH(JI), etc.  In total, there are 7*6*5… = 7! ways to shuffle the remaining athletes, and that is just for the entry that begins ABC!  We can make the same observation about entries that start ACB, BAC, etc.  So we would again deem all of these grouping “equal WRT condition #1,” and so we further revise our estimate to 10! / 3! / 7! — and that is in fact our final answer!
• In the second scenario, we want to call 1 athlete a “gold medalist,” 1 a “silver medalist,” 1 a “bronze medalist,” and 7 “non-medalists.”  Once we choose, say, Al to be our gold medalist, there is only one way to “re-shuffle” the list of entries that start with A!  We continue reasoning as above to conclude that there are now 10! / 1! / 1! / 1! / 7! outcomes.
• In the third scenario, we have 1 “gold medalist,” 3 “silver medalists,” 4 “bronze medalists,” and 2 “non-medalists.”  As above, we get 10! / 1! / 3! / 4! / 2! outcomes.

Final observation: in all three cases, each of the 10 athletes is accounted for in the computation.

• 1st situation: 10! / (3! 7!), and 3+7 = 10.
• 2nd situation: 10! / (1! 1! 1! 7!), and 1+1+1+7 = 10.
• 3rd situation: 10! / (1! 3! 4! 2!), and 1+3+4+2 = 10.

So we don’t ask about “order,” we don’t bother labeling one problem as a “combination” and another as a “permutation,” we just let common sense prevail as we ask, “how should I label each of the 10 athletes to conform to the conditions of the problem?”  Once we do this, the problem solves itself!

Note: this is the closest I get to 3-Act Math.  For the real thing, pop over to Dan Meyer’s page.

My friend Andy is getting ready to sell his house, and so he decided to undertake a few projects.  He asked me to come by and help him build a set of stairs that run from his front yard down to the back yard.  Ever think about how to build stairs? To make what you see above, we used a pre-cut template from Home Depot.  We laid the template on a piece of wood, then traced it with a pencil.  Andy used a circular saw to cut the board.  Want to see something cool?  Look at the top of the board after he made the cut: That looks about right.  Now get this: after he cuts the whole board, he flips it over…what do you think it should look like?  Think it over and then scroll down.  [Note: I made my students wait *24 hours* before I showed them this next image.]

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[spoiler space] Wow!  That shocked me to the core.  I could not believe my own eyes!  What is going on here?  [Note: I made my students wait *another 24 hours* before discussing this any further.]

So at the end of the week, I asked my students to try to explain this phenomenon of the lines-that-don’t-meet.  Suddenly I’ve got half a dozen volunteers eager to go the the board.  Students are watching each other make drawings and try to explain what’s happening.  I let each person have their say, then the next student got to go up.  Want to know the hardest part of all this?  Making the drawing.  Here are their attempts to represent the board and the saw: Wow!  It’s tough to get the picture right, isn’t it?  I think if I had a board and a saw to demonstrate with, it would be a whole lot easier.

Anyway, aside from this being a neat visual puzzler, I’m sharing this story as a testimony about *student engagement.*  Want to know the truth?  My students don’t appear to be nearly that engaged on any given day.  So here’s my question: how can we transform math class into a space where this level of engagement is the norm?  Where students are eager to go to the board and *represent their ideas with pictures?*  Where they are willing to *speak in front of their peers?*  Where they are trying their hardest to *construct a convincing argument?*  Where every person in the room *knows what the outcome looks like* but is *struggling to understand why?*

I’ve just been reading the recently published e-book Nix the Tricks.  I love the spirit of this book!  I am 100% on board.

The first “trick” the authors tackle has to do with adding integers: “keep-change-change.”  I remember hearing this for the first time when I was a brand-new teacher 11 years ago, and I was completely befuddled!  Example: 2 – -3 = 2+ +3 via keep-change-change, or so they say.

Anyway, the authors suggested remedy for this rule is to use a number line.  While that is certainly possible, I am going to take a stand here and suggest that “movement on a number line” — start here, go right or left some number of units, and end somewhere — is not the easiest approach to learning integers.  For instance — can students explain which way to go when doing 2 – -3?  Seems a bit confusing if you ask me.

I owe my preferred approach to James Tanton: just make piles of sand!  Let’s take a trip to the sandbox to see how easy it is to work with positive and negative integers. So “3” is just 3 piles of sand, and “-3” is just 3 holes in the sand.  Now what happens when I push some piles of sand into the holes? Looks like the piles got flattened!  Apparently a hole is, in some sense, “the opposite of a pile.”  So we would like to believe that 2 + -2 = 0.  This makes adding integers a piece of cake!  I’m pretty confident that this is something that children of all ages can get their minds around!  No need to worry about “moving this way or that way” on a number line.

Now for the sticky part: what about subtraction?  Well, subtraction is all fine when the numbers behave nicely: “Johnny has a group of 8 apples.  He takes away 3 of them.  How many does he have left?”  8-5 = 3 is perfectly all right for this situation, but things get a bit out of hand when we try to understand what 3-8 must mean, to say nothing of -3 – (-8)!  So what shall we do?  Just do what mathematicians do at the highest levels of the profession: ignore subtraction entirely!  From now on when we write 8 – 5, we are really saying something about *addition* — we take 8 and add the *opposite of 5,* so that 8 – 5 is — by definition — 8 + -5.  With this simple understanding, we don’t even have to talk about the following four subtraction problems, as they are merely addition problems! The only sticky point is to be careful about what we mean by -(-2).  Piece of cake!  -2 is 2 holes in the sandbox.  -(-2) is the opposite of 2 holes in the sandbox, which is………………two piles of sand!  So -(-2) really is 2 after all.

Another nice feature of this approach is that it allows us to handle algebraic expressions like the following: what is (9x+10) – (4x+3)?  Well, we really just want to *add the opposite* of 4x+3, so we have 9x+10 + -(4x+3).  What does that mean?  I want the opposite of “4x piles and 3 piles,” which is “4x holes and 3 holes,” which is -4x + -3!  So now we can write 9x + 10 + -4x + -3, and the rest is easy.

Summing up:

• Every addition problem can be handled by drawing piles and holes.
• Every subtraction problem can be construed as an addition problem.  This is not a “trick” as in “keep-change-change,” it is a sound mathematical principle: a – b = a + -b, *by definition.*
• Thus every subtraction problem can be handled via piles and holes as well, by viewing it as an addition problem!

After drawing a few million diagrams with piles and holes, children will naturally internalize the principles and begin to do these problems in their heads like grown-ups!

For more information about James Tanton’s approach, check out his newsletter where he discusses this in more detail.  And be sure to visit his web site Thinking Mathematics!