Difference between revisions of "2011 AIME II Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | <cmath>D_{1}= | + | <cmath>D_{1}=\begin{vmatrix} |
− | 10 | + | 10 |
− | \end{ | + | \end{vmatrix} = 10, \quad |
− | D_{2}= | + | D_{2}=\begin{vmatrix} |
− | + | 10 & 3 \ | |
− | 3 & 10 \ | + | 3 & 10 \ \end{vmatrix} |
− | + | =(10)(10) - (3)(3) = 91, \quad | |
− | D_{3}= | + | D_{3}=\begin{vmatrix} |
10 & 3 & 0 \ | 10 & 3 & 0 \ | ||
3 & 10 & 3 \ | 3 & 10 & 3 \ | ||
0 & 3 & 10 \ | 0 & 3 & 10 \ | ||
− | \end{ | + | \end{vmatrix}. </cmath> |
Using the expansionary/recursive definition of determinants (also stated in the problem): | Using the expansionary/recursive definition of determinants (also stated in the problem): | ||
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\end{array} } \right| = 10D_{2} - 9D_{1} = 820</math> | \end{array} } \right| = 10D_{2} - 9D_{1} = 820</math> | ||
− | This pattern repeats because the first element in the first row of <math>M_{n}</math> is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to <math>10D_{n-1}</math>. The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix <math>X_{n}</math>. <math>X_{n}</math> has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields <math>3D_{n-2} + 0(other things)=3D_{n-2}</math>. Thus, the <math>3 det(X_{n})</math> expression turns into <math>9D_{n-2}</math>. Thus, the equation <math>D_{n}=10D_{n-1}-9D_{n-2}</math> holds for all n > 2. | + | This pattern repeats because the first element in the first row of <math>M_{n}</math> is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to <math>10D_{n-1}</math>. The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix <math>X_{n}</math>. <math>X_{n}</math> has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields <math>3D_{n-2} + 0(\text{other things})=3D_{n-2}</math>. Thus, the <math>3 \det(X_{n})</math> expression turns into <math>9D_{n-2}</math>. Thus, the equation <math>D_{n}=10D_{n-1}-9D_{n-2}</math> holds for all n > 2. |
This equation can be rewritten as <math>D_{n}=10(D_{n-1}-D_{n-2}) + D_{n-2}</math>. This version of the equation involves the difference of successive terms of a recursive sequence. Calculating <math>D_{0}</math> backwards from the recursive formula and <math>D_{4}</math> from the formula yields <math>D_{0}=1, D_{4}=7381</math>. Examining the differences between successive terms, a pattern emerges. | This equation can be rewritten as <math>D_{n}=10(D_{n-1}-D_{n-2}) + D_{n-2}</math>. This version of the equation involves the difference of successive terms of a recursive sequence. Calculating <math>D_{0}</math> backwards from the recursive formula and <math>D_{4}</math> from the formula yields <math>D_{0}=1, D_{4}=7381</math>. Examining the differences between successive terms, a pattern emerges. | ||
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This is an infinite [[geometric series]] with first term <math>\frac{1}{81}</math> and common ratio <math>\frac{1}{9}</math>. Thus, the sum is <math>\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}</math>. | This is an infinite [[geometric series]] with first term <math>\frac{1}{81}</math> and common ratio <math>\frac{1}{9}</math>. Thus, the sum is <math>\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}</math>. | ||
− | Thus, <math>p + q = 1 + 72 = \boxed{073}</math> | + | Thus, <math>p + q = 1 + 72 = \boxed{073}</math>. |
+ | |||
+ | ==Solution 2 (Engineer's Induction - not rigorous)== | ||
+ | <cmath> | ||
+ | Manually, we can find | ||
+ | \[ | ||
+ | \frac{1}{8D_1+1} = \frac{1}{81}, \quad \frac{1}{8D_2+1} = \frac{1}{729}, \quad \frac{1}{8D_3+1} = \frac{1}{6561}. | ||
+ | \] | ||
+ | We notice that | ||
+ | |||
+ | Assume | ||
+ | \[ | ||
+ | \frac{\frac{1}{81}}{1 - \frac{1}{9}} = \frac{\frac{1}{81}}{\frac{8}{9}} = \frac{1}{72}. | ||
+ | \] | ||
+ | |||
+ | Thus, the answer is | ||
+ | </cmath> | ||
+ | |||
+ | ==Solution 3 (Uses first half of Solution 1)== | ||
+ | From Solution 1, | ||
+ | We can manually compute | ||
+ | |||
+ | The characteristic polynomial of this recurrence is | ||
+ | <cmath> x^2 - 10x + 9, </cmath> | ||
+ | which has roots | ||
+ | Therefore, the explicit formula for | ||
+ | <cmath> D_n = \alpha_1 + 9^n \alpha_2. </cmath> | ||
+ | |||
+ | We can solve for | ||
+ | <cmath> \alpha_1 + \alpha_2 = 1, </cmath> | ||
+ | <cmath> \alpha_1 + 9\alpha_2 = 10. </cmath> | ||
+ | |||
+ | Solving these equations, we find | ||
+ | <cmath> \alpha_1 = -\frac{1}{8}, \quad \alpha_2 = \frac{9}{8}. </cmath> | ||
+ | |||
+ | Substituting these back into the explicit formula, we find | ||
+ | <cmath> D_n = \frac{9^{n+1} - 1}{8}. </cmath> | ||
+ | |||
+ | Notice that | ||
+ | <cmath> \frac{1}{8D_{n+1} + 1} = 9^{-(n+1)}. </cmath> | ||
+ | |||
+ | The summation asks us to find the sum of a geometric series with ratio | ||
+ | The sum of this series is | ||
+ | <cmath> \frac{\frac{1}{81}}{1 - \frac{1}{9}} = \frac{\frac{1}{81}}{\frac{8}{9}} = \frac{1}{72}. </cmath> | ||
+ | |||
+ | Thus, the answer is | ||
+ | <math>\boxed{73}</math> | ||
==See also== | ==See also== | ||
− | {{AIME box|year=2011|n=II|num-b= | + | {{AIME box|year=2011|n=II|num-b=10|num-a=12}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:16, 25 November 2024
Contents
[hide]Problem
Let be the matrix with entries as follows: for , ; for , ; all other entries in are zero. Let be the determinant of matrix . Then can be represented as , where and are relatively prime positive integers. Find .
Note: The determinant of the matrix is , and the determinant of the matrix ; for , the determinant of an matrix with first row or first column is equal to , where is the determinant of the matrix formed by eliminating the row and column containing .
Solution
Using the expansionary/recursive definition of determinants (also stated in the problem):
This pattern repeats because the first element in the first row of is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to . The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix . has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields . Thus, the expression turns into . Thus, the equation holds for all n > 2.
This equation can be rewritten as . This version of the equation involves the difference of successive terms of a recursive sequence. Calculating backwards from the recursive formula and from the formula yields . Examining the differences between successive terms, a pattern emerges. , , , , . Thus, .
Thus, the desired sum is
This is an infinite geometric series with first term and common ratio . Thus, the sum is .
Thus, .
Solution 2 (Engineer's Induction - not rigorous)
Solution 3 (Uses first half of Solution 1)
From Solution 1,
The characteristic polynomial of this recurrence is
which has roots
We can solve for
Solving these equations, we find
Substituting these back into the explicit formula, we find
Notice that
The summation asks us to find the sum of a geometric series with ratio
Thus, the answer is
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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