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, an+1=49nan=49n25(9n+(1)n+1)=49n259n25(1)n+1=(425)9n25(1)n+2(1)=1859n+25(1)n+2=25(9n+1+(1)n+2), as desired. $\square$

Solution by TheUltimate123.

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See also

2018 USAJMO (ProblemsResources)
First Problem Followed by
Problem 2
1 2 3 4 5 6
All USAJMO Problems and Solutions