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2018 USAJMO Problems/Problem 1

Revision as of 02:08, 22 February 2019 by Aopsal (talk | contribs)

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.

Solution 3

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 $a_2 = 32$.

From here, we proceed by complementary counting. We first count the total amount of numbers that satisfy both bullets but that may have zero as its first digit. The units digit can be one of four primes, and each digit to the left will have 9 choices (any digit but the one that was just used). Then the total for this group of numbers is just $4*9^{n-1}$.

Now we must subtract all numbers in the above group that have 0 as its first digit. This is just $a_{n-1}$ because for each number belonging to that group, we could add a zero to the front to achieve a number in the first group. Then we have the recursive formula $a_n = 4*9^{n-1}-a_{n-1}$.

To simplify this, we take a look at the first few terms. We see that

\[a_2 = 4*9^1-a_1 = 36-4 = 4(9-1) = 4(9^1-9^0)\] \[a_3 = 4*9^2-a_2 = 4(81) - 4(9-1) = 4(9^2-9^1+9^0)\]

We see a pattern where $a_n = 4(9^{n-1}-9^{n-2}+9^{n+3}...+(-1)^{n-1}9^0)$ and we can prove that it holds for all $a_n$ because subtracting a_{n-1} from 4(9^n) is the same thing as reversing all previous signs of the preceding powers of 9. This constitutes an alternating pattern, which we can calculate as a geometric series. The first term is $4(9^{n-1})$ and the common ratio is $-\frac{1}{9}$, so 

\[a_n = \frac{4(9^{n-1})(1-(-\frac{1}{9})^n)}{1-(-\frac{1}{9})} = \frac{4(9^{n-1}-(-1)^n(-\frac{1}{9)}{\frac{10}{9}}\] (Error compiling LaTeX. Unknown error_msg)

I am working on it right now

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

See also

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