Birthday Number Puzzle VariationsDate: 11/18/2002 at 15:40:08 From: Jeanette Birkland Underwood Subject: Number Trick Dear Dr. Math, A friend of mine recently showed me the following trick: 1. First of all, pick the number of times a week that you would like to have chocolate. (Try for more than once but fewer than 10 times.) 2. Multiply this number by 2 (just to be bold). 3. Add 5. (for Sunday) 4. Multiply it by 50 - I'll wait while you get the calculator... 5. If you have already had your birthday this year add 1752... If you haven't, add 1751... 6. Now subtract the four-digit year that you were born. 7. You should have a three-digit number. The first digit is your original number (i.e., how many times you want to have chocolate each week). And the next two numbers are your age! I tried it, and it works no matter what number I start with. How does it work? He also said that this year (2002) is the only year it will ever work. Is that true? Thanks, -J Date: 11/18/2002 at 16:11:05 From: Doctor Ian Subject: Re: Number Trick Hi Jeanette, Probably the best way to see why this works is to avoid committing to a specific number and see what happens. Instead of picking a number, I'm going to use 'n' to represent some choice I haven't yet made. Let's see what happens as I follow the steps: 1) n 2) 2n 3) 2n + 5 4) 100n + 250 5) 100n + (2001 or 2002) 6) 100n + (2001 or 2002) - (year of birth) Now, let's look at what I ended up with. If we write it this way, 100n + (2001 or 2002) - (year of birth) = 100n + [(year of my last birthday) - (year of my birth)] = 100n + [my age] This might be a good time to note that if you start with a number greater than 9, or if you're more than 99 years old, the trick won't work. So basically, the trick is saying that 50(2n + 5) + 1750 + i - (year of birth) where i (for 'increment') is the number of years past 2000 for the year when you had your last birthday, equal to 100n + (your age) which is, of course, true (for 0 < n < 10, and age < 100): 50(2n + 5) + 1750 + i - (yob) = (100n + 250) + 1750 + i - (yob) = 100n + (250 + 1750 + i) - yob = 100n + (2000 + i) - (yob) = 100n + (your age) It _is_ true that with these specific numbers, the trick will work only this year. However, if you want it to work next year, you can just change i to '2 or 3' instead of '1 or 2'. Some other variations on the problem would be 25(4n + 10) + 1750 + i - (year of birth) 20(5n + 10) + 1800 + i - (year of birth) 4(25n + 127) + 1492 + i - (year of birth) 4(25n + 224) + 1776 + i - (year of birth) In general, you have a(bn + c) + d + i - (year of birth) and the constraints you have to meet are 1) ab = 100 2) ac + d = 2000 Let's say I want to make d = 1908, the last year the Chicago Cubs won the World Series. Then ac + 1908 = 2000 ac = 2000 - 1908 ac = 92 = 2 * 2 * 23 The possible values of a and c are: a = 1, c = 92 a = 2, c = 46 a = 4, c = 23 a = 23, c = 4 a = 96, c = 1 But 2 and 4 are the only values of a that divide 100 evenly, so I can have either 2(50n + 46) + 1908 + i - (year of birth) or 4(25n + 23) + 1908 + i - (year of birth) - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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