Difference between revisions of "2018 USAJMO Problems/Problem 1"
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{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>a_n</math> be the number of <math>n</math>-digit numbers that satisfy the condition that most-left digit different than <math>0</math>. Let <math>b_n</math> be the number of <math>n</math>-digit numbers that satisfy the condition that most-left digit is <math>0</math>. We can write that | ||
+ | <cmath> | ||
+ | \left\{ | ||
+ | \begin{split} | ||
+ | a_n + b_n & = 4\cdot 9^{n-1} \\ | ||
+ | b_n & = a_{n-1} | ||
+ | \end{split} | ||
+ | \right. | ||
+ | </cmath> | ||
+ | and so we find that | ||
+ | <cmath> a_n + a_{n-1} = 4\cdot 9^{n-1} \tag{1}</cmath> | ||
+ | <math>x+1</math> is the characteristic polynomial of the homogeneous recurrence relation <math>a_n + a_{n-1} = 0 </math>. Then <math>x=-1</math> is a root. Also, by the non-homogeneous part of <math>4\cdot 9^{n-1}</math>, we understand that <math>x=9</math> is another root of the recurrence relation <math>(1)</math> with the complementary solution idea. Therefore, <math>(a_n)</math> sequence in the form <cmath> a_n =C_19^n + C_2(-1)^n .</cmath> | ||
+ | We can calculate that <math>C_1 = \dfrac{2}{5}, C_2 = - \dfrac{2}{5}</math> with using <math>a_1=4, a_2=32</math>. Thus we get: | ||
+ | <cmath> a_n = \dfrac{2}{5}\left(9^{n} - (-1)^{n} \right) </cmath> | ||
+ | |||
+ | Lokman GÖKÇE | ||
+ | |||
==Video Solution== | ==Video Solution== |
Revision as of 11:34, 28 February 2023
Contents
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.
Solution 5
Let be the number of
-digit numbers that satisfy the condition that most-left digit different than
. Let
be the number of
-digit numbers that satisfy the condition that most-left digit is
. We can write that
and so we find that
is the characteristic polynomial of the homogeneous recurrence relation
. Then
is a root. Also, by the non-homogeneous part of
, we understand that
is another root of the recurrence relation
with the complementary solution idea. Therefore,
sequence in the form
We can calculate that
with using
. Thus we get:
Lokman GÖKÇE
Video Solution
https://www.youtube.com/watch?v=1BijQXkC5dg -Osman Nal -inserted by srisainandan6
See also
2018 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |