Difference between revisions of "2018 USAJMO Problems/Problem 1"
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Solution by putnam-lowell. | Solution by putnam-lowell. | ||
+ | ==Solution 3== | ||
+ | The answer is <math>\boxed{\frac{2}{5}\left(9^n-(-1)^{n}\right)}</math>. | ||
+ | |||
+ | As in the first solution, let <math>a_n</math> denote the number of <math>n</math>-digit numbers that satisfy the condition. Clearly <math>a_1 = 4</math> and <math>a_2 = 32</math>. | ||
+ | |||
+ | 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 <math>4*9^{n-1}</math>. | ||
+ | |||
+ | Now we must subtract all numbers in the above group that have 0 as its first digit. This is just <math>a_{n-1}</math> 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 <math>a_n = 4*9^{n-1}-a_{n-1}</math>. | ||
+ | |||
+ | To simplify this, we take a look at the first few terms. We see that | ||
+ | |||
+ | <cmath>a_2 = 4*9^1-a_1 = 36-4 = 4(9-1) = 4(9^1-9^0)</cmath> | ||
+ | <cmath>a_3 = 4*9^2-a_2 = 4(81) - 4(9-1) = 4(9^2-9^1+9^0)</cmath> | ||
+ | |||
+ | We see a pattern where <math>a_n = 4(9^{n-1}-9^{n-2}+9^{n+3}...+(-1)^{n-1}9^0)</math> and we can prove that it holds for all <math>a_n</math> 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 <math>4(9^{n-1})</math> and the common ratio is <math>-\frac{1}{9}</math>, so | ||
+ | |||
+ | <cmath>a_n = \frac{4(9^{n-1})(1-(-\frac{1}{9})^n)}{1-(-\frac{1}{9})}</cmath> | ||
+ | |||
+ | I am working on it right now | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:05, 22 February 2019
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 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
.
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 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
and the common ratio is
, so
I am working on it right now
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 |