Ordering Games

Ordering games involve estimation and comparison, two essential skills for mathematics.

Here are some ordering games I gave to groups of students at Greenwich High School.  Teams raced to arrange the slips of paper from smallest to largest.  (No calculator)

Some groups were very enthusiastic about this game; others not so much.  In the future it might help to do one “practice” round with numbers like 1,2,3,4,5, so that each group could get a feel for what they are trying to do and what a successful ordering looks like.

Each ordering game below is geared to a different skill level.  The cards are not necessarily in order as pictured.

A little bit of trig:

Finding areas enclosed by curves.  I gave this to students in a calculus class, but it could also be great for eager students who haven’t learned calculus yet.  It can be completed without any calculus.

I feel I have only scratched the surface of what can be done with ordering games.  It would be cool to create decks of cards (several different decks) like this.  You’d need an ordering game of 13 items.  The concept of suits (clubs, diamonds, hearts, spades) would be preserved.  Just as in a normal deck, there would be 4 copies of every number.  However, math skills would be required to sort the cards.  For instance, instead of relationships like “a jack is higher than a 10” you’d have relationships like “a card with log(11) is higher than a card with sin(89 degrees).”  Students could then play classic card games with each other, using the mathematical card decks.  They’d have to swap out decks periodically to avoid getting too familiar with one deck and memorizing the orderings.

Eventually students could design decks for each other.  This would involve thinking of 13 numbers that are obscure enough to make people think, but “easy” enough that no laborious calculations would be needed to compare them to one another.

Exposed Industrial Ceilings

Exposed ceilings are really hip these days, creating an industrial warehouse feel in office spaces and restaurants.  They don’t cover up the wiring and beams, so you can see what’s going on.

How is this related to math?

Well, try to simplify the fraction 275/11.  If you’re not sure where to start, use the Exposed Ceiling Technique.  Break everything down so you can see the metaphorical beams:

275/11 = (25*11)/11

Now it’s easy to notice the 11’s and cancel them out.

This is also useful when solving problems.

Often you’ll be plugging in values and you’ll want to multiply them out into one big number.  Resist this temptation.  Don’t rush to plaster things over.  Show your exposed ceilings some love and they will love you back.

Suppose you’re plugging values into an equation and you end up with,

(3*11*12*x)/(34) = (121)/(17)

Using the Exposed Ceiling Technique, i.e. without multiplying or calculating anything, we have:

x = (121*34)/(3*11*12*17)

…instead of rushing to multiply, expose it further…

x = (11*12*17*2)/(3*11*12*17)

Now we can eliminate 11, 12, and 17:

x = 2/3.

Exposed ceilings enable you to see patterns that would otherwise be obscured.

Suppose you are exploring the concept of compound interest for the first time.  You are trying to derive a formula.  You come across a problem:

\$2000 is invested in a savings account with 1% interest, compounded annually.  How much is in the account after 5 years?

After some thought, you decide that you need to repeat a process each year: write down the amount you had at the beginning of the year, and take 101% percent of it to find out how much you have at the end of the year.  Taking 101% of a number is the same as multiplying it by 1.01.

Let’s see how this works.

First, let’s do this without exposed ceilings:

starting amount: \$2000

after 1 year: \$2000*(1.01) = \$2020 <– this is \$2000 plus an extra 1% of \$2000

after 2 years: \$2020*(1.01) = \$2040.20

after 3 years: \$2040.20*(1.01) = \$2060.60

after 4 years: \$2060.60*(1.01) = \$2081.21

after 5 years: \$2081.21*(1.01) = \$2102.02

final amount: \$2102.02

You were able to solve the problem, but no obvious formula popped out of this work.

Now let’s do this with exposed ceilings:

This time, don’t multiply anything out until the end.  This is also a convenient approach if you don’t have a calculator.

starting amount: \$2000

after 1 year: \$2000*(1.01) <– this is \$2000 plus an extra 1% of \$2000

after 2 years: \$2000*(1.01)*(1.01)

after 3 years: \$2000*(1.01)*(1.01)*(1.01) = \$2000*(1.01)3

after 4 years: \$2000*(1.01)*(1.01)*(1.01)*(1.01) = \$2000*(1.01)4

after 5 years: \$2000*(1.01)*(1.01)*(1.01)*(1.01)*(1.01) = \$2000*(1.01)5

final amount: ok, now you can multiply it out. \$2000*(1.01)= \$2102.02

You got the same answer, but this time you also exposed an underlying pattern: to find the amount after the nth year, just take the original amount, and multiply it by 1.01n.  In other words,

final amount after n years = (starting amount)*(1.01)n

Not only did your exposed ceilings save you from boring extra calculations, they gave you a glimpse at the underlying structure, which enabled you to write a general formula.