1990 USAMO Problems/Problem 1
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 |