Difference between revisions of "1990 USAMO Problems/Problem 1"
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[[Category:Olympiad Combinatorics Problems]] | [[Category:Olympiad Combinatorics Problems]] |
Revision as of 13:57, 21 October 2007
Problem
A license plate has six digits from 0 to 9 and may have leading zeros. If two plates must always differ in at least two places, what is the largest number of plates that is possible?
Solution
We show by induction that we can find a set of plates for digits. It is true for : take the plates . Suppose it is true for . If d is a digit from to and is a plate of digits, let be the plate of digits which has a as its first digit, and the remaining digits the same as those of , except that the last digit is that for plus (reduced if necessary). Let be a set of plates for digits. We claim that the set $S' = { [d, s] : d = 0, 1, ...$ (Error compiling LaTeX. Unknown error_msg) or and belongs to } is a set of plates for digits. It obviously has times as many members as , so this claim is sufficient to establish the induction.
We have to show that and differ in at least two places. If , then , so s and t differ in at least two places. The same change is made to their last digits, so and differ in at least two places. If $a ≠ b$ (Error compiling LaTeX. Unknown error_msg) and , then and differ in both their first and last places. If and , then and differ in at least two places and so the modified s and t, differ in at least one place. But and also differ in the first place, so they differ in at least two places.
So we have established that the largest number is at least for digits.
But any two plates which differ only in the last digit cannot both be chosen. So at most of the possible plates can be chosen. That shows that is best possible.
See also
1990 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |