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Birthday Number Puzzle Variations

Date: 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/ 
Associated Topics:
Middle School Algebra
Middle School Puzzles

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