Unbelievable math problem

[Bofinn]

Here is a math trick so unbelievable that it will stump you.

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?

Cool huh........

Have a Great Day!!! -

[FB]

weird.

[PCBruiser]

Well, not so magical at all. Let's look at the math:

Assume y = first three digits of your phone number, and x = last 4 digits. The formula then becomes:

250(80y+1)+2x-250
_________________ = 10,000y+x.
2

The factor of 10,000 just moves the position of y four places to the left (producing y0000), and then you just add x. Since you have supplied both y and x, and since phone numbers are always 7 digits, this will always produce your phone number. In essence, there's nothing magical or unbelieveable about it at all, just simple math.

[Joe Public]

Phone numbers in Norway are 8 digits, but it came close.

[PCBruiser]

For an 8 digit phone number just set y to equal the first 4 digits and it will work exactly. Or for a 6 digit phone number set y to the first 2 digits, and it will still work.

[Bofinn]

Originally Posted by PCBruiser

Well, not so magical at all. Let's look at the math:

I knew somebody in here would be smart enough to figure it out! WTG PCB!

ps don't let KT know you are a smart dog, he might pee on you.

[Drufuss]

Originally Posted by PCBruiser

Well, not so magical at all. Let's look at the math:

Assume y = first three digits of your phone number, and x = last 4 digits. The formula then becomes:

250(80y+1)+2x-250
_________________ = 10,000y+x.
2

The factor of 10,000 just moves the position of y four places to the left (producing y0000), and then you just add x. Since you have supplied both y and x, and since phone numbers are always 7 digits, this will always produce your phone number. In essence, there's nothing magical or unbelieveable about it at all, just simple math.

Puff, the magic has gone.

Leave it to the mods to spoil all the fun.

Greetings

Dru

[KingTermite]

Originally Posted by Bofinn

ps don't let KT know you are a smart dog, he might pee on you.

I think I need to pee on YOU for that!

[polonyc2]

Originally Posted by PCBruiser

Well, not so magical at all. Let's look at the math:

Assume y = first three digits of your phone number, and x = last 4 digits. The formula then becomes:

250(80y+1)+2x-250
_________________ = 10,000y+x.
2

The factor of 10,000 just moves the position of y four places to the left (producing y0000), and then you just add x. Since you have supplied both y and x, and since phone numbers are always 7 digits, this will always produce your phone number. In essence, there's nothing magical or unbelieveable about it at all, just simple math.

I'm more impressed with PCB's explanation of the solution then the original problem

PCB musta aced his Trig and Calculus classes...

[k0NG0]

Still, factoring integers is more fun. Do it in Z[i] (the gaussian integers) and it's even more fun. Algebra in general is a hoot, especially the abstract kind.

[PCBruiser]

Originally Posted by polonyc2

I'm more impressed with PCB's explanation of the solution then the original problem

PCB musta aced his Trig and Calculus classes...

Yes, I did, but that's what a degree in Math from MIT does for you.

[KingTermite]

Originally Posted by kONGO

Still, factoring integers is more fun. Do it in Z[i] (the gaussian integers) and it's even more fun. Algebra in general is a hoot, especially the abstract kind.

I guess I'm more visual......really loved Trig and Analytical Geometry type stuff better.

Although...I do have a special place in my heart for differential equations (especially LaPlace Transforms).

[Centurion]

Hey guys,

I just saw this thread, and so I thought I'd ask about a math problem that I am struggling with. Basically, I am wanting to know how to calculate the value of a number that has an exponent which is a complex number (real + imaginary number).

For example: 5^(3+5i) = ?

[grsamf]

There is a much more complicated little problem that uses birthdate and address in that order. When you get done, you wind up with address first and then the birthdate. Again, it is basically simple math.

On the other hand, PCB, people thought there was something mystical and magical about this and you burst the bubble. Do you go to department stores at Christmas and tell the lttle kids there is no Santa, too?

[k0NG0]

5^(3+5i) = (5^3)*(5^5i) first of all.

Then write 5 = e^(ln(5)) and you get 125*((e^(ln(5))^5i) which is 125*e^(5i*ln(5)) which is 125*e^(i*5*ln(5)).

Then remember Euler's e^ix = cos(x) + i*sin(x) and you get: 125*(cos(5*ln(5)) + i*sin(5*ln(5))).

The a = e^(ln(a)) is the key here. It's a nice little trick to pull, I think.