Difference between revisions of "2018 USAJMO Problems/Problem 1"
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Solution by aopsal | Solution by aopsal | ||
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+ | ==Solution 4== | ||
+ | Let <math>f(n)</math> denote the number of <math>n</math>-digit positive integers satisfying the conditions listed in the problem. | ||
+ | Claim 1: <math>f(n) = 8f(n-1) + 9f(n-2). </math> | ||
+ | To prove this, let <math>k</math> be the leftmost digit of the <math>n</math>-digit positive integer. When <math>k</math> ranges from <math>1</math> to <math>9, </math> the allowable second-to-leftmost digits is the set <math>\{0, 1, 2, …, 9\}, </math> with <math>k</math> excluded. Note that since <math>1, 2, …, 9</math> are all repeated <math>8</math> times and using our definition of <math>f(n)</math> (since <math>n</math>-digit positive integers must begin with a positive digit), we have proved that <math>8f(n-1)</math> part. Since <math>0</math> is repeated <math>9</math> times, and since digits that are allowed to be adjacent to the <math>0</math> (the third-to-leftmost digit of our positive integer) are integers ranging from <math>1</math> to <math>9, </math> we have proved the <math>9f(n-2)</math> part similarly. We can see that <math>\frac{f(1)}{4} = 1</math> and <math>\frac{f(2)}{4} = 8, </math> by simple computation. These will be our base cases. | ||
+ | Now, let <math>s(n) = \frac{f(n)}{4}.</math> | ||
+ | Claim 2: <math>s(n+1) = 9s(n) + (-1)^n</math> | ||
+ | We can prove this using induction. Let <math>s(a) = k</math> and <math>s(a+1) = 9k - 1. </math> By plugging <math>k</math> and <math>9k - 1</math> into our formula <math>s(a+2) = 8s(a+1) + 9s(a), </math> we obtain <cmath>s(a+2) = 8(9k-1) + 9k \implies s(a+2) = 81k - 8 = 9s(a+1) + 1.</cmath>Now, by plugging <math>9k - 1</math> and <math>81k - 8</math> into the same formula, we obtain <math>s(a+3) = 8(81k - 8) + 9(9k - 1) = 729k - 73 = 9s(a+2) - 1. </math> Since with <math>a+2</math> and <math>a+3</math> we’re back where we started (this is since <math>s(a+3) = 9s(a+2) - 1</math> just like <math>s(a+1) = 9s(a) - 1</math>), and using our base cases of <math>s(1) = 1</math> and <math>s(2) = 8, </math> our induction is complete. | ||
+ | Claim 3: <math>s(n) = \frac{9^n + (-1)^{n+1}}{10}</math> | ||
+ | We can prove that <cmath>s(n) = \underbrace{9(9(9(…9(9-1)+1)…-1)+1)-1}_{\text{n-1 nested parentheses}} = \sum_{i=0}^{n-1} 9^{i} (-1)^{n-i-1}</cmath>by simply using our base case of <math>s(1) = 1</math> and repeatedly using the distributive property to obtain further values of <math>s(n). </math> For the case where <math>n</math> is even, we can manipulate the summation into <math>\sum_{i=0}^{\frac{n}{2} - 1} 8 \cdot 9^{2i}</math> which equals <math>\frac{8 \cdot 9^{n} - 1}{9^2 - 1} = \frac{9^n - 1}{10}, </math> by geometric series. For the case where <math>n</math> is odd, we can manipulate the summation into <math>1 + \sum_{i=0}^{\frac{n-1}{2} - 1} 8 \cdot 9^{2i + 1}</math> which equals <cmath>9 \cdot \frac{8 \cdot 9^{n-1} - 1}{80} + 1 = \frac{9^n - 9}{10} + 1 = \frac{9^n + 1}{10}.</cmath>Thus, we have proved Claim 3 for both odd and even <math>n.</math> | ||
+ | |||
+ | Since <math>f(n) = 4s(n),</math> our answer is <math>4 \cdot \frac{9^n + (-1)^{n+1}}{10} = \boxed{ \frac{2}{5} (9^n + (-1)^{n+1})}. </math> | ||
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+ | -fidgetboss_4000 | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:46, 16 June 2020
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 . 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.
Solution 2
The answer is .
As in the first solution, let denote the number of -digit numbers that satisfy the condition. Clearly and . We claim that for we have the recurrence .
To prove this, we split the -digit numbers satisfying the conditions into cases depending on whether or not the second digit is . If the second digit is nonzero, our number is formed from one of the numbers with one fewer digit satisfying the conditions, times possible choices of adding a digit to the left. If the second digit is zero, our number is formed from one of the numbers with two fewer digits satisfying the conditions, times possible choices of adding 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 , where the are constants to be determined from the initial conditions and , and and are the roots of the corresponding quadratic equation . We solve and get , so our roots are and . Now, we use our conditions and to derive the system of linear equations
Solving this system yields and , and we are done.
Solution by putnam-lowell.
Solution 3
The answer is .
As in the first solution, let denote the number of -digit numbers that satisfy the condition. Clearly and .
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 .
Now we must subtract all numbers in the above group that have 0 as its first digit. This is just because for each number in the first group beginning with a zero, we could take away the zero, leaving us with a number that works for the case (this is true because the next number would not be zero, or the consecutive digit requirement would be violated). Then we have the recursive formula .
To simplify this, we take a look at the first few terms. We see that
We see a pattern where and we can prove that it holds for all because subtracting from 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 and the common ratio is , so
We are done.
Solution by aopsal
Solution 4
Let denote the number of -digit positive integers satisfying the conditions listed in the problem. Claim 1: To prove this, let be the leftmost digit of the -digit positive integer. When ranges from to the allowable second-to-leftmost digits is the set with excluded. Note that since are all repeated times and using our definition of (since -digit positive integers must begin with a positive digit), we have proved that part. Since is repeated times, and since digits that are allowed to be adjacent to the (the third-to-leftmost digit of our positive integer) are integers ranging from to we have proved the part similarly. We can see that and by simple computation. These will be our base cases. Now, let Claim 2: We can prove this using induction. Let and By plugging and into our formula we obtain Now, by plugging and into the same formula, we obtain Since with and we’re back where we started (this is since just like ), and using our base cases of and our induction is complete. Claim 3: We can prove that by simply using our base case of and repeatedly using the distributive property to obtain further values of For the case where is even, we can manipulate the summation into which equals by geometric series. For the case where is odd, we can manipulate the summation into which equals Thus, we have proved Claim 3 for both odd and even
Since our answer is
-fidgetboss_4000
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 |