Math Problem - Not Quite Humor

[Johnny]

This is not really humor though some might think it's funny. It's just one of those things that fascinates me. As it says, use a calculator.

Here is a math trick so unbelievable that it will stump you.
Personally I would like to know who came up with this and why that
person is not running the country or at least the treasury department.

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 ?

 

[TallPaul]

It works out algebracially to

10,000X + Y
where X is the first three digits and Y is the last 4 digits.

You'll see that 80x250 is 20,000 then it gets divided by 2 later. You'll also see that the 250x1 is subtracted out and the last 4 digit number is entered twice because it also gets divided by two.

Now that I have been such a sourpuss to spoil the trick, and it is a fun one at that, try this on your calculator:

Enter 1134. Invert calculator. Read entry.

 

[drivewaytech]

Of course there is the calculator trick from back when I was a kid. I can't remember exactly how it was worded, but went something like this:

Enter the numbers as they appear, no addition or subtraction. If Suzie weighed 553 pounds, ran 7800 miles non-stop in 8 days, what would she be.... (turn the calculator upside down)

 

[Quattro Pete]

Originally Posted By: Johnny
Here is a math trick so unbelievable that it will stump you.

Hmm... it asks you for your phone number and then it gives it back to you. Unbelievable?

 

[TallPaul]

Another math trick that is interesting is e to the (Pi times i) = -1. The fascinating aspect of this to me is that it puts all the very special numbers (e, Pi, i, 1, and 0) in one equation:

e to the (Pi times i) +1 = 0

 

[crinkles]

the OP trick only works on 7 digit phone mumbers. ours are 8 digit dumbers, so yuo lose the 4th digit

 

[crinkles]

Originally Posted By: TallPaul
Another math trick that is interesting is e to the (Pi times i) = -1. The fascinating aspect of this to me is that it puts all the very special numbers (e, Pi, i, 1, and 0) in one equation:

e to the (Pi times i) +1 = 0


AKA the most elegant mathematical formula

 

[MolaKule]

You may have seen this one before, but take the first three integers and double them as in 113355.

Now separate the first three from the last three: 113 355

Now divide the last set of integers by the first

355/113

 

[digitalSniperX1]

...except when i = 0...but I'm assuming i is the imaginary operator and equals (-1)^1/2.

For all integers k, e^(k*i*pi) are equal to -1 (except when k=0).

Also akin to and more generalized with Euler's identity...

e(jwt) = cos(wt)+jsin(wt)

(this equation is uses as part of a great many expression in radar and could be considered the most fundamental equation in radar signal processing...something discussed in another section lately...only to be closed for some unknown reason).


wt can be any irrational real multiplicand of PI..or the set of real numbers from + to - infinity and j is the imaginary operator (electrical engineers use j, not i, since i is used in equations to represent current). Of course the integer portion of wt/2pi would be the number revolutions around the unit circle...because you see...

this when plotted in the complex plane results in a unit circle.

Applying a basic trig identity:

1 = (cos(wt)^2 + (jsin(wt))^2)^1/2

Fairly basic stuff...although I could be wrong....and in that case this is just a joke.

 

[TallPaul]

Originally Posted By: MolaKule
You may have seen this one before, but take the first three integers and double them as in 113355.

Now separate the first three from the last three: 113 355

Now divide the last set of integers by the first

355/113
That is really neat. Wonder how accurate it is.

===========================================================
As for 8 digit phone numbers, the trick could be revised to accomodate, but not worth the bother IMO.

 

[MolaKule]

Quote:
You may have seen this one before, but take the first three integers and double them as in 113355.


Should have read, "You may have seen this one before, but take the first three integers and repeat them as in 113355."

Use 64 bit precision or the highest precision your calc. or PC will go.

 

[UncleS2]

Originally Posted By: MolaKule
Quote:
You may have seen this one before, but take the first three integers and double them as in 113355.


Should have read, "You may have seen this one before, but take the first three integers and repeat them as in 113355."

Use 64 bit precision or the highest precision your calc. or PC will go.


Yup, you can get a good approximation, accurate to about as many places as you'd like, or ever make practical use of. But it will *Never* be exactly Pi.

(Why not? It may be as accurate as your hardware/software can utilize, but you still have a rational number, ie integer/integer. Pi is an irrational number. Done.)


BTW- I've had 3 different people email me the "phone #" gimmick in the past few years. I've shown them all how it works, step by step, and how it can easily be modified(say, add a few steps & include the area code). They've all called me a party pooper! Ah, the curses of having a logical mind- and a degree in everyone's favorite subject, Math.

 

[TallPaul]

I discovered that the pop up calculator on my work computer goes out 32 decimal places. Anyway, Pi to anything beyond 2 digits is more than I'll ever need for any thing practical. But for pure math pursuits, one can never get enough digits of Pi.

 

[jaj]

Originally Posted By: TallPaul
I discovered that the pop up calculator on my work computer goes out 32 decimal places. Anyway, Pi to anything beyond 2 digits is more than I'll ever need for any thing practical. But for pure math pursuits, one can never get enough digits of Pi.


My kids and stepkids all know pi to 9 places of decimal (they find it useful, lets just say). I memorized it in university to 17 places and can rattle it off today with no problem.

Mola's calculation is good to 6 places of decimal with the 7th ok for rounding but incorrect otherwise.

 

[tropic]

Originally Posted By: TallPaul
...try this on your calculator:

Enter 1134. Invert calculator. Read entry.

When I was in elementary school, the magic number was 5318008

 

[TallPaul]

tropic, sounds like the description of some of the students (that can be taken more than one way I guess).

Another common approx of Pi is 22/7. There are others (obviously).

 

[crinkles]

Originally Posted By: jaj
Originally Posted By: TallPaul
I discovered that the pop up calculator on my work computer goes out 32 decimal places. Anyway, Pi to anything beyond 2 digits is more than I'll ever need for any thing practical. But for pure math pursuits, one can never get enough digits of Pi.


My kids and stepkids all know pi to 9 places of decimal (they find it useful, lets just say). I memorized it in university to 17 places and can rattle it off today with no problem.

Mola's calculation is good to 6 places of decimal with the 7th ok for rounding but incorrect otherwise.


closest i'll ever need in engineering is 3.14159

even that is too accurate given that the error margin on natural materials (soil and dirt) are sometimes order of magnitudes.

 

[TallPaul]

Originally Posted By: crinkles

closest i'll ever need in engineering is 3.14159

even that is too accurate given that the error margin on natural materials (soil and dirt) are sometimes order of magnitudes.
Presumably you are a civil engineer. Is it possible that a mechanical engineer may need a couple more digits?

 

[Kestas]

Good enough for any engineer. Mathemeticians are a different story.

 

[Shannow]

Saw an absolute revelation the other day.

And am now going to sit down and do it.

have a co-ordinate box, 0,0 to 1,1...area 1 square units.

Throw a huge number of random co-ordinates into the box (I don't think that they need to be random, but that was the premise of it).

Take each and every co-ordinate, and calculate the square root of the (X^2xY^2), i.e the distance from 0,0.

If it's less than, it goes in one bucket. Greater than, the other bucket. The "exactly" 1 gets distributed in the ratio of the buckets.

The first bucket is an approximation to pi/4.

More random numbers (or IMO the tighter the grid), the closer you are.