Difference between revisions of "Mock AIME II 2012 Problems/Problem 15"
(Created page with "==Problem:== Define <math>a_n=\sum_{i=0}^{n}f(i)</math> for <math>n\ge 0</math> and <math>a_n=0</math>. Given that <math>f(x)</math> is a polynomial, and <math>a_1, a_2+1, a_3+8...") |
|||
Line 4: | Line 4: | ||
==Solution:== | ==Solution:== | ||
'''Lemma:''' <math>f(x)</math> is a quadratic | '''Lemma:''' <math>f(x)</math> is a quadratic | ||
+ | |||
+ | |||
'''Proof: ''' Note that using the method of finite differences, once we get to a constant term, the rest of the terms in the polynomial that have not been eliminated are going to be <math>0</math>. | '''Proof: ''' Note that using the method of finite differences, once we get to a constant term, the rest of the terms in the polynomial that have not been eliminated are going to be <math>0</math>. | ||
Since <math>a_1, a_2+1, a_3+8, a_4+27, a_5+64, a_6+125</math> is an arithmetic sequence, difference(let this be h)<math>=a_2-a_1+1=a_3-a_2+7=a_4-a_3+19=a_5-a_4+37=a_6-a_5+61</math>, we get in general <math>\sum_{i=0}^{n+1}f(i)-\sum_{i=0}^{n}f(i)=a_{n+1}-a_n=f(n+1)</math>. Therefore <math>f(2)+1=f(3)+7=f(4)+19=f(5)+37=f(6)+61</math>. We now have equations with our polynomials. Subtract all consecutive equations to give us <math>f(3)-f(2)=-6</math>, <math>f(4)-f(3)=-12</math>, <math>f(5)-f(4)=-18, f(6)-f(5)=-24</math>. Let <math>f(x)=b_n*x^n+b_{n-1}*x^{n-1}+b_{n-2}*x^{n-2}+\cdots b_2*x^2+b_1*x+b_0</math>. Note that subtracting two equations eliminates <math>a_0</math>, and we are going to be taking two more differences to get the equations equal to <math>0</math>. These two more differences subtract the <math>b_1</math> term and the <math>b_2</math> term, because by method of finite differences, you only have to take <math>1</math> and <math>2</math> differences respectively to eliminate the linear/quadratic term. Therefore <math>f(x)</math> is quadratic. <math>\blacksquare</math> | Since <math>a_1, a_2+1, a_3+8, a_4+27, a_5+64, a_6+125</math> is an arithmetic sequence, difference(let this be h)<math>=a_2-a_1+1=a_3-a_2+7=a_4-a_3+19=a_5-a_4+37=a_6-a_5+61</math>, we get in general <math>\sum_{i=0}^{n+1}f(i)-\sum_{i=0}^{n}f(i)=a_{n+1}-a_n=f(n+1)</math>. Therefore <math>f(2)+1=f(3)+7=f(4)+19=f(5)+37=f(6)+61</math>. We now have equations with our polynomials. Subtract all consecutive equations to give us <math>f(3)-f(2)=-6</math>, <math>f(4)-f(3)=-12</math>, <math>f(5)-f(4)=-18, f(6)-f(5)=-24</math>. Let <math>f(x)=b_n*x^n+b_{n-1}*x^{n-1}+b_{n-2}*x^{n-2}+\cdots b_2*x^2+b_1*x+b_0</math>. Note that subtracting two equations eliminates <math>a_0</math>, and we are going to be taking two more differences to get the equations equal to <math>0</math>. These two more differences subtract the <math>b_1</math> term and the <math>b_2</math> term, because by method of finite differences, you only have to take <math>1</math> and <math>2</math> differences respectively to eliminate the linear/quadratic term. Therefore <math>f(x)</math> is quadratic. <math>\blacksquare</math> |
Revision as of 02:27, 5 April 2012
Problem:
Define for and . Given that is a polynomial, and is an arithmetic sequence, find the smallest positive integer value of such that .
Solution:
Lemma: is a quadratic
Proof: Note that using the method of finite differences, once we get to a constant term, the rest of the terms in the polynomial that have not been eliminated are going to be .
Since is an arithmetic sequence, difference(let this be h), we get in general . Therefore . We now have equations with our polynomials. Subtract all consecutive equations to give us , , . Let . Note that subtracting two equations eliminates , and we are going to be taking two more differences to get the equations equal to . These two more differences subtract the term and the term, because by method of finite differences, you only have to take and differences respectively to eliminate the linear/quadratic term. Therefore is quadratic.
Now, let . Since , we have or . Since , we get and . Subtract the LHS from the RHS of the equations to give us and . Subtract these two equations to give us or . Now, substitute this into to give us or . Therefore .
Since , we get , we are going to have this being true for . Since we want being positive, we use $\negative$ (Error compiling LaTeX. Unknown error_msg) for to give us . The RHS is the same as . Our goal now is to find approximately. Note that (where , and are digits of ), therefore , and substitute this into our equation for to give a smallest possible value of being to give us and hence the smallest possible positive integer value for is .