2018 USAJMO Problems/Problem 1
Problem
For each positive integer , find the number of
-digit positive integers that satisfy both of the following conditions:
no two consecutive digits are equal, and
the last digit is a prime.
Solution 1
The answer is .
Suppose denotes the number of
-digit numbers that satisfy the condition. We claim
, with
.
[i]Proof.[/i] It is trivial to show that
. Now, we can do casework on whether or not the tens digit of the
-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in
ways and choose the units digit in
ways, since it must be prime and not equal to the tens digit. Therefore, there are
ways in this case.
If the tens digit is not prime, we can use complementary counting. First we consider the number of -digit integers that do not have consecutive digits. There are
ways to choose the first digit and
ways to choose the remaining digits. Thus, there are
integers that satisfy this. Therefore, the number of those
-digit integers whose units digit is not prime is
. It is easy to see that there are
ways to choose the units digit, so there are
numbers in this case. It follows that
and our claim has been proven.
Then, we can use induction to show that . It is easy to see that our base case is true, as
. Then,
which is equal to
as desired.
Solution by TheUltimate123.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2018 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |