Trying to Unlock a Cell Phone

Date: 11/16/2003 at 12:45:46
From: Savannah
Subject: four number combination to a locked cellular phone

I put in the local newspaper that I found a cellular phone on the side
of the road.  No one answered the ad so now I thought I would use it
but it's locked by a four digit number.  I've tried a lot of
combinations and can't seem to get the right one.  Can you help?

Savannah

Date: 01/05/2005 at 14:52:55
From: Doctor Douglas
Subject: Re: four number combination to a locked cellular phone

Hi Savannah.

Although we can't tell you what the code is to unlock the phone, we
can compute the number of possible combinations (thus giving you some
information about how long the task might take).

The number of possibilities for the initial digit is 10, because the
initial digit could be any of {0,1,2,3,4,5,6,7,8,9}.  For exactly the 
same reason, the number of possibilities for the second digit is 10.  
This means that the number of possible combinations for *just the 
first two* digits is

  10 x 10 = 100.

In fact, it's not too hard to list the possible sets of two 
consecutive digits.  Here's the list, where I use dots (...) to 
save me from writing out every element of this somewhat boring 
list: {00,01,02,03,...,08,09,10,11,...,18,19,20,21,22,...,97,98,99}.  
You can see that these are just the numbers from zero to 99, where 
we include leading zeroes.  

Now let's add the third digit.  There are ten possibilities here, 
and these combine with each of the hundred possibilities for the
first two digits:  {000,001,002,...,997,998,999}.  There are 
10 x 10 x 10 = 1000 combinations in this list.  You can probably 
see the pattern now:  for a sequence of N symbols, each of which
has K possibilities, the number of possible combinations is

    K x K x ...x K    =    K^N
     \__________/
     there are N           "K to the power N"
     factors of K
    in this product

You can apply this formula to calculate the number of four-digit
PIN combinations.

This formula is a direct result of the "Fundamental Principle of
Counting", which can be paraphrased as "If there are P ways of doing
one thing and Q ways of doing another, then there are P x Q ways of
doing both".
 
I should mention that although I have used the word "combinations"
here (in keeping with its usage with respect to locks and locking),
the word "combination", as used by most mathematicians, has a 
different meaning:  a way of choosing a subset of some elements from
a larger set, without regard to ordering or sequence.  For example,
choosing a set of three representatives from a group of fifteen.
For more information about these sorts of possibilities, you can
check out the following web page:

  Ask Dr. Math FAQ:  Permutations and Combinations
    http://mathforum.org/dr.math/faq/faq.comb.perm.html 

In your problem above, each of the four elements IS distinct from
each other (i.e. knowing the first digit of the code is completely
independent from knowing the last digit of the code).  Your problem 
can be thought of in these terms as a "permutation with replacement",
as illustrated by the following answer in our archives:

  Combinations and Permutations with Replacement
    http://mathforum.org/library/drmath/view/64626.html 

Our archives also have many problems that illustrate the use of the
Fundamental Principle of Counting (counting-by-multiplication):

  Combinations of Three Words
    http://mathforum.org/library/drmath/view/56148.html 

  Path Possibilities
    http://mathforum.org/library/drmath/view/56208.html 

  Maximum Possible Combinations
    http://mathforum.org/library/drmath/view/56130.html 

  Combinations of Images on a TV Screen
    http://mathforum.org/library/drmath/view/56108.html 

  Counting Answer Keys
    http://mathforum.org/library/drmath/view/56112.html 

  Examples of the Fundamental Counting Principle
    http://mathforum.org/library/drmath/view/54298.html 

I hope that this helps you with your cell phone.  If you decide that 
there are too many combinations to try each one individually, I 
suggest that you take the phone to one of the service providers that 
use that phone.  They usually have special equipment that can find 
out to whom the phone is registered, or if it is your own phone, to 
"reset" it so that it is usable.
 
- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/