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To: kormac who wrote (38338)	10/5/1998 11:59:00 AM
From: Lee Bush	Respond to of 41046

seppo: Can you please work out a game whereby VALSPEC will disappear. Thank you, Lee

To: kormac who wrote (38338)	10/5/1998 12:34:00 PM
From: VALUESPEC	Read Replies (2) \| Respond to of 41046

Seppo, where your math is correct and your reasoning is outstanding, I think the following is a more lucid way of looking at the problem: (2x + 5) 50 + 1748 - y = 100x + z Where: x= the days dining out y= the year of your birth z= your age By simplifying the problem, much understanding can be obtained: 100x + 250 + 1748 - y = 100x + z 100x + 1998 - y = 100x + z As in our method, you can now see that 100 times x (dinning days) makes your single digit choice a three digit number (i.e. 1 day is made into 100). Since most ages are under 100, for most ages, the first digit will the days dining. The next two will be your age. For instance, 168 would be 1 day dinning (1 x 100 put it to the left !), 68 years. However, if you are 100 years old, the result will be 200. As you can see by looking at my simplification, Seth's game really boils down to: 100x + 1998 -y = 100x + y Dining days per week times 100, plus 1998, minus your age. In this light, the problem is much easier to understand. All they did was break up 1998 into 5*50 + 1748 (or 1747 depending on if your date passed yet). I might add that my equation is fair and easy to use even though it has three variables because you already know x and y, x being the your preference for dining and y your own age ! VALUESPEC