Associated Topics || Dr. Math Home || Search Dr. Math

### 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).

4. Multiply it by 50 - I'll wait while you get the
calculator...

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

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

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)

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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search