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3,5,12

-- BernieJohnston - 09 Oct 2005

Yup, that's what we got, EXCEPT we spent some time trying to make the x-cubed equation come out to 1800, not 1880, because I mis-copied it.

-- CatherineJohnson - 09 Oct 2005

I wonder when we're getting the hay baler problem?

-- CatherineJohnson - 09 Oct 2005

So is this the portion of the mathematics school year when they get megahard problems one after the other?

-- CarolynJohnston - 09 Oct 2005

you mean this is homework for a public school?
wow. very impressive if any goodly portion
of the students can get an answer.

(assume one has $x, y, z \in \Natural_Numbers$
-- we will modify this if it should become necessary.)

i myself took the bait and thought about
the "first" two conditions first. for example,
just about the first thing i looked at was
"okay; then the biggest cube is less than
(or equal to) 1880 ... (fiddle, fiddle) ...
12^3 = 1728 ; 13^3 = 2197 .... right!
the biggest of x, y, & z is less or equal to 12."
(step 1)

somewhere in here i noticed that either two
or none of x, y, z were odd ...
(one does this almost unconsciously after a while ...
the various applications of "even or odd"
are called "parity" arguments [and may very well be
the best-possible motivating examples
for "modular arithmetic"]). but, heck, that doesn't
seem to be all that useful either.

finally i hit on an idea that worked.
xyz = 180 is, in hindsight, the right place to start.
factor 'er: 180 = (2^2)(3^2)(5) is the "factored form"
(known to exist uniquely by the so-called
"fundamental theorem of arithmetic").
now it's just a matter of fiddling around with
2, 2, 3, 3, and 5 until we can form the x, y, z we want.
if i'd've had any sense, i'd've prob'ly tried
"12" first (in light of the upper bound
in step 1) ... but no; i had to try
4*15*3, 10*6*3, and 10*9*2 first. but when i saw
12*3*5 fall out, i instantly saw that x+y+z was right;
it remained only to check that the sum of cubes
was also correct. yep. fun!

-- VlorbikDotCom - 09 Oct 2005

what did christopher do?

-- VlorbikDotCom - 09 Oct 2005

Again, this is more puzzle than math, especiallt when you consider your typical middle school's understanding of math. Better to give three difficult linear equations that can be solved computationally.

-- KDeRosa - 09 Oct 2005

all--

So is this the portion of the mathematics school year when they get megahard problems one after the other?

Carolyn's referring to something she and I have talked about a lot.

Last year the Phase 4 class was a source of huge conflict between parents & the administration.

This year the course seems quite a bit improved.

One of the problems last year was that the teacher spent the first month on problem solving, which meant the kinds of problems I've just posted, organized into the standard categories: guess and check, make a list, work backwards, and so on. Basically, bowdlerizations of Polya.

This year the teacher started with chapter one in the textbook, so I thought the Extended Problems might have been axed.

But it seems what they've done is to distribute them throughout the course, rather than chunk them into one unit.

back to Carolyn: I'll see what happens. They just finished chapter one, and took their test.

I think that went well. Christopher got a 90, and the class average was 88--and the test looks suitably demanding to me. So the fact that they all did so well seems very good; these kids may be acquiring mastery (or preliminary mastery....)

I'm not sure whether she's moving on to Chapter 2 or not.

We'll see.

-- CatherineJohnson - 09 Oct 2005

V

you mean this is homework for a public school?
wow. very impressive if any goodly portion
of the students can get an answer.

I'd be stunned if there is a single child in the class who can do any part of these questions without help from his parent.

-- CatherineJohnson - 09 Oct 2005

V

Wow!

I love your account!

(WE HAVE TO ARCHIVE THIS STUFF!)

It would be interesting to know.......what this means.

I now, having been exposed to Extended Problems enough to begin to form an Extended Problem Plan, instantly look for the hand grip.

I spent zero time messing around with the X cubed equation, zero.

And I planned to spend zero time.

I could see that xyz was the hand grip, and I started there.

As a matter of fact, Christopher hasn't worked the problem yet; I put that up front because if anyone at the school were to read that post I don't want them to think Christopher got the answer from the brainiacs at Kitchen Table Math, which he didn't, and won't.

(Of course, someone from the school could always read this Comment thread!)

I've been mulling over how to get Christopher to do as much of the problem as possible on his own, and I think I've come up with an idea.

I'm going to print out the list of Middle School Math topics from edhelper.com, and give it to Christopher, and say, "These are the topics you either know now, or will be learning soon. Which one of these might work to get a start on this problem?"

Then I hope he's going to pick prime factoring.

If he doesn't, I'm going to prompt him to pick it.

I'll also ask him which of the 3 equations looks easiest to start with.

I'm going to take notes, too, and report back.

-- CatherineJohnson - 09 Oct 2005

I actually found this problem simple, mainly because I've been seeing so much of this stuff - but also because I've been doing so much elementary mathematics. HUGE amounts of elementary mathematics.

I'm probably developing expertise specifically in elementary mathematics.

Nothing beyond it, but elementary mathematics: I'm getting to have a big fat Domain of procedural, declarative, & conceptual knowledge of grade school arithmetic.

-- CatherineJohnson - 09 Oct 2005

Another thing: last year Christopher had a horrifically difficult find-a-pattern Extended Problem sent home. I couldn't even begin to do it; no one could. (It turned out that the pattern was related to the letter of the alphabet each number stood for.)

I called my neighbor, who bails me out on a lot of these things, and she said, instantly, 'Factoring almost always works with this stuff.'

That is now my First Step in approaching Challenge Problems.

Prime factor the sucker.

-- CatherineJohnson - 09 Oct 2005

Alistair (oldest stepson) commented that the math SAT really consisted of a handful of tricks and a little domain knowledge, applied over and over.

You have to know about the Pythagorean theorem, and 30-60-90 and 45-45-90 triangles and the ratios of their sides, for example; that crops up over and over. There are a few other things like that that the math SAT focuses on.

Really, to do well on the math SAT requires domain knowledge in a limited subset of middle and high school level math.

Although I understand there have been some major changes in the SAT in the last couple of years. Does anyone know anything about that?

-- CarolynJohnston - 09 Oct 2005

Apparently they're going to put some algebra II questions on.

The Princeton Review series is brilliant; that's the book to get.

-- CatherineJohnson - 09 Oct 2005

Initial report: I tried showing Christopher the list of middle school topics, and it was a bomb.

Didn't help at all.

-- CatherineJohnson - 09 Oct 2005