janet_rose
Posted : 7/8/2008 7:18:56 AM
sandra_slayton
Just how in the blazes does this work???????
Assume you have already had your birthday. The algebraic formula for this "puzzle" would be
[ (2x + 5 ) * 50 ] + 1758 - y
where x is the times you want to eat out and y is the year you were born. This works out to be
100x + 250 + 1758 - y = 100x + ( 2008 - y )
( 2008 - y ) will be your two digit age if you have already had your birthday this year. However, that assumes that you are less than 100 years old.
Changing 1758 to 1757 just subtracts 1 when you haven't had your birthday yet this year.
If you want this to work next year, use 1759/1758 instead of 1758/1757. That will give you ( 2009 - y ) for folks that have already had their birthday.
In fact the numbers to use in any year would be (the current year minus 250)/(the current year minus 249).