2018 USAJMO Problems/Problem 1

Problem

For each positive integer $n$ , find the number of $n$ -digit positive integers that satisfy both of the following conditions:

$\bullet$ no two consecutive digits are equal, and

$\bullet$ the last digit is a prime.

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 $a_1=4$ . $\textit{Proof.}$ 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, $a_{n+1}=4\cdot 9^n-a_n=4\cdot 9^n-\frac{2}{5}\left(9^n+(-1)^{n+1}\right)=4\cdot 9^n-\frac{2}{5}\cdot 9^n-\frac{2}{5}(-1)^{n+1},$ which is equal to $a_{n+1}=\left(4-\frac{2}{5}\right)\cdot 9^n-\frac{2}{5}\cdot\frac{(-1)^{n+2}}{(-1)}=\frac{18}{5}\cdot 9^n+\frac{2}{5}\cdot(-1)^{n+2}=\frac{2}{5}\left(9^{n+1}+(-1)^{n+2}\right),$ as desired. $\square$

Solution by TheUltimate123.

Solution 2

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

As in the first solution, let $a_n$ denote the number of $n$ -digit numbers that satisfy the condition. Clearly $a_1 = 4$ and $a_2 = 32$ . We claim that for $n \geq 3$ we have the recurrence $a_n = 8a_{n-1} + 9a_{n-2}$ .

To prove this, we split the $n$ -digit numbers satisfying the conditions into cases depending on whether or not the second digit is $0$ . If the second digit is nonzero, our number is formed from one of the $a_{n-1}$ numbers with one fewer digit satisfying the conditions, times $8$ possible choices of adding a digit to the left. If the second digit is zero, our number is formed from one of the $a_{n-2}$ numbers with two fewer digits satisfying the conditions, times $9$ possible choices of adding $0$ and then any nonzero digit to the left. This proves our claim.

This gives us a linear three-term recurrence. It is well-known that its solution is of the form $c_1r_1^n + c_2r_2^n$ , where the $c_i$ are constants to be determined from the initial conditions $a_1$ and $a_2$ , and $r_1$ and $r_2$ are the roots of the corresponding quadratic equation $x^2 = 8x + 9$ . We solve and get $(x-9)(x+1) = 0$ , so our roots are $r_1 = 9$ and $r_2 = -1$ . Now, we use our conditions $a_1 = 4$ and $a_2 = 32$ to derive the system of linear equations

$9c_1 - c_2 = 4$ $81c_1 + c_2 = 32$

Solving this system yields $c_1 = \frac{2}{5}$ and $c_2 = -\frac{2}{5}$ , and we are done.

Solution by putnam-lowell.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2018 USAJMO Problems/Problem 1

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Problem

Solution 1

Solution 2

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