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m00n

Math trick

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1. Grab a calculator. (you won't be able to do this one in your head)

2. Key in the first three digits of your phone number (NOT the area code)

3. Multiply by 80

4. Add 1

5. Multiply by 250

6. Add the last 4 digits of your phone number

7. Add the last 4 digits of your phone number again.

8. Subtract 250

9. Divide number by 2

Do you recognize the answer?

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Nope!

It must be a mystery number, or my old girlfriends address. Either way don't recognize it.

Moon maybe the formula you gave us is wrong[;)]

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Glad you guys liked it, but I want to make sure you guys understand, I didn't make that formula. That whole trick was sent to me via email. I'm just passing it along. [:)]

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I was going to ask what kind of people have the time to develop these formulas, chain letters, etc.?. It really amazes me, because a lot of time has to go into a formula like that, or maybe it just comes easy to some people?

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This would be a cute algebra problem for students to express as an

equation. After you express the equation and reduce it, all

you're doing is multiplying the first three digits of your phone number

by 10,000 (moving the decimal point four places to the right) and then

adding the last four digits of your phone number. It just looks

like magic because of the obtuse mathematical pathway.

This is how you would express it:

[(80X+1)250 + Y + Y - 250]1/2 = your phone number

(20,000X + 250 + 2Y - 250)1/2 = your phone number

or

10,000X + Y = your phone number

where:

X = your phone number prefix, and

Y = the last four digits of your phone number

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I liked this m00n. It's the kind of thing that could spark a kids interest in math. Thanks for posting.

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I liked this m00n. It's the kind of thing

that could spark a kids interest in math. Thanks for posting.

or make him hate his teachers for giving him such an "annoying"

question [;)] But it really is a good skill to learn how to break down

seemingly complex alogorithms into very simple equations (that's what

writing algorithms is all about). I've got a really good textbook here

(well my dad's old textbook as he was way more into this stuff that I

will ever be) that gets into some really crazy alogithm logic and the

limitations of math (like algorithms that are completely true, but

mathematically can't be - which means the math fails). I can only

remember part of the books title: "Escher Bach and M_____" (some dude

who's name that starts with M). Yes, Bach had a lot to do with

alogrithm writing and that's why I like the book so much (because there

are many many references to music). I'm not sure if it was this book,

but there are theories that we can define all of music with a single

algorithm and anything that doesn't fit is just heard as noise (btw,

this is completely different than the algorithms driving Bach's music).

But I

think we can save these topics for another day [;)]

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Who,

I'll bet the book that you're thinking of is "Goedel, Escher, and Bach; An Eternal Golden Braid" by Douglas Hofstader. In his mid twenties, Kurt Goedel (the "dude whose name starts with M") published his "Incompleteness Theorems" proving that for an axiomatic system of mathematics, there are problems that cannot be proven to be true or false within mathematical axioms as had been previously assumed. Simply put, his proof demonstrated that math, in itself, is not complete or perfect and implied that mathematics could not fully be expressed by algorithm. His work turned some math theorists on their ears (including Bertand Russell and David Hilbert).

Speaking of algorithms, here is another cool math problem that I pulled from a biography of Richard Feynman:

You are paddling a canoe up a river. The river is flowing at a rate 3 miles per hour. You can paddle your canoe at a constant rate of 4.5 miles per hour. At some point, your hat blows off and lands in the water and floats down stream with the current. You keep paddling upstream but 45 minutes after your hat blew into the river, you decide to turn the canoe around (instantaneously and without acceleration) and go get your hat. At this point, how long will it take you to go back and get your hat?

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Ya that's the one....I need to go read it again as I don't think my 12

year old mind was able to comprehend it fully (though I still think

that might be the case now, lol). Dunno how I got M and G mixed up

though! Must be another one of his books....he's got tons of them.

So we gonna get into special relativity now with the canoe analogy?

Frame of references are fun...my favorite is that you can fit a 100

yard long rocket travelling near the speed of light inside a 100 foot

long barn [:D] (and there are experiments where they do this too). Now

if only we could find a way to shut the doors of the barn and make them

strong enough to withstand the impace of the "rocket" - now that would

be something real interesting to do....

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Nice catch on the canoe problem. I really like the forehead slap most people give themselves after getting tangled up in a bunch of 3's, 4's, 1/3's and 1/4's

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I did it the hard way first. Pick the spot on the shore. The canoe goes north for 45 minutes to get 1.125 miles. The hat goes south to get 2.25 miles. So at turn around they are 3.375 miles apart. Then the guy has to paddle 3.375 miles at 4.5 miles per hour, which is 0.75 hours, or 45 minutes.

That gives a hint to the easy answer. The river is the frame of reference. The hat can't paddle and is stationary. So if you paddle away at some speed (any constant speed) for X time, it will take you X time again to travel back at that speed.

- - - There are two math teachers on a road 20 miles apart. There is fly on the handlebars of one. . . . . .

Gil

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Well done Gil. Here is another problem from Richard Feynman that he used to start arguments among physicists at parties.

It's obvious that the sprinkler nozzles below will rotate clockwise when water flows out of the nozzles under pressure. If you submerge the sprinkler in a tank of water and suck water into the sprinkler (reversing the flow), which way will the nozzles rotate?

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I don't see any problem with it turning backwards. We could look at it a couple of ways.

When there is water flow out this is because the pressure at the nozzle hole is higher than the surrounding water. If we add up all the pressure squares on the sprinkler as a whole, this means there is force left over which moves it up to some clockwise speed where the drag of the surround water balances the forces.

If there is flow in, this is because the pressure at the nozzle hole is lower than the surrounding water. If we add up all the pressure squares on the sprinkler as a whole, this means there is force left over which moves it up to some anti-clockwise speed where the drag of the water balances the forces.

I'll take a page from D-Man's book, i.e. his observation that speakers suck half the time. None the less, the overall position of the box does not change so average velocity, and average acceleration, must be zero. If we rapidly reverse the flow of the water in the sprinker head, it should average to zero change of position (rotation). If we know it goes clockwise half the time, it must go anti-clockwise the other half the time.

I think he doesn't mean to introduce the problem that eventually the water in the pool will turn.

Gil

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