# Re: [Fwd: Fw: Freaky Formula]

• Subject: Re: [amsat-bb] [Fwd: Fw: Freaky Formula]
• From: "Kenneth J. Ernandes" <n2wwd@xxxxxxxxxxxxxx>
• Date: Tue, 07 Jul 1998 18:58:57 -0400

```The formula is not at all freaky.

1.  Multiplying the number in step #1 by two and then later by 50 is
the same as multiplying it by 100 -- hence the first digit.
2.  Multiplying 5 by 50 and is 250.  Adding 1748 to 250 is 1998 (this year)
then he year of your last birthday is 1997 = 250 + 1747.
3.  Subtracting your birth year from the year of your last birthday will
give your age -- hence the last two digits.

BTW, the formula will work next year if you use 1749 instead of 1748 and
1748 instead of 1747 in step #5.

Is it obvious in hindsight?

73, Ken N2WWD
n2wwd@amsat.org
http://www.mindspring.com/~n2wwd

At 15:30 7/7/98 -0400, you wrote:
>>This really works...try it...Have a great 4th of July.
>>>IT  REALLY WORKS. DON'T CHEAT BY SCROLLING DOWN FIRST!!!
>>>It only takes 30 seconds so work this out as you read.
>>>Don't read the bottom until you have worked it out!!!
>>>
>>>1.  First of all, pick the number of days a week that you would like
>>>     to go out for dinner.
>>>
>>>2.  Multiply this number by 2.
>>>
>>>
>>>4.  Multiply it by 50.
>>>
>>>     If you haven't, add 1747.
>>>
>>>6.  Last step: Subtract the four digit year that you were born
>>>     (e.g. 1958). see below:
>>>
>>>
>>>
>>>RESULTS:
>>>
>>>You should now have a three digit number:
>>>
>>>The first digit of this was your original number (i.e. how many times you
>>>want
>>>to go out each week).
>>>
>>>The second two digits are your age!!!
>>>
>>>Freaky, huh?
>>>.
>>>This is the only year it will ever work,
>
>>>
>>>
>>>
>>>
>>>
>>>
>>
>
Ken Ernandes
n2wwd@amsat.org
http://www.mindspring.com/~n2wwd

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```

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