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Ehsan
November 23, 2003, 01:21 PM
The follwing calculation is for a tel number in Canada and US, don't know about other countries if it works. Got this in a forward e-mail, this thing really works.

1) Key in the first 3 digits of your phone
number into a calculator (do not key in your area code).
2) Multiply by 80.
3) Add 1.
4) Multiply by 250.
5) Plus last 4 digits of your phone number.
6) Plus last 4 digits of your phone number again.
7) Minus 250.
8) Divide by 2.

Shubho
November 25, 2003, 10:24 PM
hahahah, ehsan, man, that was hilarious.

a much simpler version of that would be to do the following:
1. Multiply first 3 digits by 10,000
2. Add last 4 digits

This is algebraically equivalent to what you just did.

Proof:

Let x = 'first 3 digits of telephone number'
Let y = 'last 4 digits of telephone number'

The steps as proposed by ehsan:
1. 80x
2. 80x + 1
3. (80x + 1)250
4. (80x + 1)250 + y
5. (80x + 1)250 + 2y
6. (80x + 1)250 + 2y - 250
7. ((80x + 1)250 + 2y - 250)0.5

Simplification of the above:
((80x + 1)250 + 2y - 250)0.5 = (20,000x + 250 + 2y - 250)0.5

= (20,000x + 2y)0.5

= 10,000x + y

The fact that I went through the trouble to post this shows that I am a lame idiot who has nothing better to do on Eid day. :P

Orpheus
November 25, 2003, 11:47 PM
this thing really works

Bravo! Ehsan Bravo! and your major is Engineering? :D

Good job Shubho on deriving the fomula... You are the fermat of Banglacricket!

chinaman
November 26, 2003, 12:53 AM
Better yet and of course without multiplication, write all the digits of your phone number from left to right in correct order. Now read the whole string. See?
Not bad, huh? Want to know how I was able to come up with the easiest formula? Well, I was the very best student in math of all the 4th graders in my school! Now top that.

Carte Blanche
November 26, 2003, 02:25 AM
hahahahaha chinaman