# 2018 USAJMO Problems/Problem 1

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## Problem

For each positive integer $n$, find the number of $n$-digit positive integers that satisfy both of the following conditions: [list] [*] no two consecutive digits are equal, and [*] the last digit is a prime. [/list]

## Solution 1

The answer is $\boxed{\frac{2}{5}\left(9^n+(-1)^{n+1}\right)}$.

Suppose $a_n$ denotes the number of $n$-digit numbers that satisfy the condition. We claim $a_n=4\cdot 9^{n-1}-a_{n-1}$, with $a_1=4$. [i]Proof.[/i] It is trivial to show that $a_1=4$. Now, we can do casework on whether or not the tens digit of the $n$-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in $a_{n-1}$ ways and choose the units digit in $3$ ways, since it must be prime and not equal to the tens digit. Therefore, there are $3a_{n-1}$ ways in this case.

If the tens digit is not prime, we can use complementary counting. First we consider the number of $(n-1)$-digit integers that do not have consecutive digits. There are $9$ ways to choose the first digit and $9$ ways to choose the remaining digits. Thus, there are $9^{n-1}$ integers that satisfy this. Therefore, the number of those $(n-1)$-digit integers whose units digit is not prime is $9^{n-1}-a_{n-1}$. It is easy to see that there are $4$ ways to choose the units digit, so there are $4\left(9^{n-1}-a_{n-1}\right)$ numbers in this case. It follows that $$a_n=3a_{n-1}+4\left(9^{n-1}-a_{n-1}\right)=4\cdot 9^{n-1}-a_{n-1},$$ and our claim has been proven.

Then, we can use induction to show that $a_n=\frac{2}{5}\left(9^n+(-1)^{n+1}\right)$. It is easy to see that our base case is true, as $a_1=4$. Then, \begin{align*} a_{n+1}&=4\cdot 9^n-a_n\\ &=4\cdot 9^n-\frac{2}{5}\left(9^n+(-1)^{n+1}\right)\\ &=4\cdot 9^n-\frac{2}{5}\cdot 9^n-\frac{2}{5}(-1)^{n+1}\\ &=\left(4-\frac{2}{5}\right)\cdot 9^n-\frac{2}{5}\cdot\frac{(-1)^{n+2}}{(-1)}\\ &=\frac{18}{5}\cdot 9^n+\frac{2}{5}\cdot(-1)^{n+2}\\ &=\frac{2}{5}\left(9^{n+1}+(-1)^{n+2}\right), \end{align*} as desired. $\square$

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 byProblem 2 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions
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