Solution to Problem 2
Congratulations and $5.00 go to this week’s winner:
Kevin Bourrillion
There were nine answers received. Correct solutions were also received from: Kevin Kerr, Mike
Fitzpatrick, Joshua Beenders, Joe Faust, and John Dahlstrom.
The answer is: The registrar can issue 10,000 IDs.
The following explanation is courtesy of this week’s winner. Let’s call two numbers
confusable
if
they differ in only one digit.
We first observe that there can be
at most
10,000 possible IDs issued. From any block of ten
consecutive numbers sharing the first four digits, for example, from the numbers 61610 to 61619, you
can assign at most one number, since any two numbers from this block would differ in only the last
digit, making them confusable. T here are 10,000 such blocks of ten numbers, which gives the upper
bound of 10,000 IDs.
To finish the job we have to come up with a way of choosing the final digit from each of the 10,000
blocks of ten digits so as to avoid confusable numbers. The easiest way to do this is:
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 Spring '10
 AlbertoDelgado
 Combinatorics, Englishlanguage films, Summation, The Easiest Way, Mike Fitzpatrick

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