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Difference between revisions of "Mock AIME II 2012 Problems/Problem 15"
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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, a_4+27, a_5+64, a_6+125, \cdots</math> is an arithmetic sequence, find the smallest positive integer value of <math>x</math> such that <math>f(x)<-2012</math>.   
 
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, a_4+27, a_5+64, a_6+125, \cdots</math> is an arithmetic sequence, find the smallest positive integer value of <math>x</math> such that <math>f(x)<-2012</math>.   
  
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==Solution:==
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==Solution 1==
 
'''Lemma:'''  <math>f(x)</math> is a quadratic
 
'''Lemma:'''  <math>f(x)</math> is a quadratic
  
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Since <math>f(x)<-2012</math>, we get <math>-3x^2+9x+2012<0</math>, we are going to have this being true for <math>x> \frac{-9\pm \sqrt{9^2-4(-3)(2012)}}{-6}</math>.  Since we want <math>x</math> being positive, we use <math>\negative</math> for <math>\pm</math> to give us <math>x>\frac{-9-\sqrt{24225}}{-6}</math>.  The RHS is the same as <math>\frac{9+\sqrt{24225}}{6}</math>.  Our goal now is to find <math>\sqrt{24225}</math> approximately.  Note that <math>155^2=\underline{15*16},2,5=24025</math> (where <math>15*16, 2</math>, and <math>5</math> are digits of <math>155^2</math>), therefore <math>156>\sqrt{24225}>155</math>, and substitute this into our equation for <math>x</math> to give a smallest possible value of <math>x</math> being <math>\frac{156+9}{6}>x>\frac{155+9}{6}</math> to give us <math>27.5>x>27.\overline{3}</math> and hence the smallest possible positive integer value for <math>x</math> is <math>\boxed{028}</math>.
 
Since <math>f(x)<-2012</math>, we get <math>-3x^2+9x+2012<0</math>, we are going to have this being true for <math>x> \frac{-9\pm \sqrt{9^2-4(-3)(2012)}}{-6}</math>.  Since we want <math>x</math> being positive, we use <math>\negative</math> for <math>\pm</math> to give us <math>x>\frac{-9-\sqrt{24225}}{-6}</math>.  The RHS is the same as <math>\frac{9+\sqrt{24225}}{6}</math>.  Our goal now is to find <math>\sqrt{24225}</math> approximately.  Note that <math>155^2=\underline{15*16},2,5=24025</math> (where <math>15*16, 2</math>, and <math>5</math> are digits of <math>155^2</math>), therefore <math>156>\sqrt{24225}>155</math>, and substitute this into our equation for <math>x</math> to give a smallest possible value of <math>x</math> being <math>\frac{156+9}{6}>x>\frac{155+9}{6}</math> to give us <math>27.5>x>27.\overline{3}</math> and hence the smallest possible positive integer value for <math>x</math> is <math>\boxed{028}</math>.
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 +
==Solution 2==
 +
Another way of finding <math>f(x)</math> is noticing that <math>a_n-a_{n-1}=\sum_{i=0}^{n}f(i)-\sum_{i=0}^{n-1}f(i)= f(n)</math> and that the terms of the arithmetic sequence are <math>a_n+(n-1)^2</math>. The difference of arithmetic sequences are constant, so <math>a_n+(n-1)^2-(a_{n-1}+(n-2)^2)= f(n) + 3n^2 - 9n + 7</math> is constant. Let that constant be <math>c</math>. Then <math>f(n)+3n^2-9n+7 =c</math> and so <math>f(n) = -3n^2+9n +7+c</math>. The solution follows as above.

Revision as of 16:37, 27 December 2012

Problem:

Define $a_n=\sum_{i=0}^{n}f(i)$ for $n\ge 0$ and $a_n=0$. Given that $f(x)$ is a polynomial, and $a_1, a_2+1, a_3+8, a_4+27, a_5+64, a_6+125, \cdots$ is an arithmetic sequence, find the smallest positive integer value of $x$ such that $f(x)<-2012$.

Solution 1

Lemma: $f(x)$ 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 $0$. Since $a_1, a_2+1, a_3+8, a_4+27, a_5+64, a_6+125$ is an arithmetic sequence, difference(let this be h)$=a_2-a_1+1=a_3-a_2+7=a_4-a_3+19=a_5-a_4+37=a_6-a_5+61$, we get in general $\sum_{i=0}^{n+1}f(i)-\sum_{i=0}^{n}f(i)=a_{n+1}-a_n=f(n+1)$. Therefore $f(2)+1=f(3)+7=f(4)+19=f(5)+37=f(6)+61$. We now have equations with our polynomials. Subtract all consecutive equations to give us $f(3)-f(2)=-6$, $f(4)-f(3)=-12$, $f(5)-f(4)=-18, f(6)-f(5)=-24$. Let $f(x)=b_n*x^n+b_{n-1}*x^{n-1}+b_{n-2}*x^{n-2}+\cdots b_2*x^2+b_1*x+b_0$. Note that subtracting two equations eliminates $a_0$, and we are going to be taking two more differences to get the equations equal to $0$. These two more differences subtract the $b_1$ term and the $b_2$ term, because by method of finite differences, you only have to take $1$ and $2$ differences respectively to eliminate the linear/quadratic term. Therefore $f(x)$ is quadratic. $\blacksquare$

Now, let $f(x)=b_2x^2+b_1x+b_0$. Since $a_n=0$, we have $\sum_{i=0}^{0}f(0)=0$ or $b_0=0$. Since $f(2)+1=f(3)+7=f(4)+19$, we get $4b_2+2b_1+1=9b_2+3b_1+7$ and $9b_2+3b_1+7=16b_2+4b_1+19$. Subtract the LHS from the RHS of the equations to give us $5b_2+b_1+6=0$ and $7b_2+b_1+12=0$. Subtract these two equations to give us $2b_2+6=0$ or $b_2=-3$. Now, substitute this into $5b_2+b_1+6=0$ to give us $-15+6+b_1=0$ or $b_1=9$. Therefore $f(x)=-3x^2+9x$.

Since $f(x)<-2012$, we get $-3x^2+9x+2012<0$, we are going to have this being true for $x> \frac{-9\pm \sqrt{9^2-4(-3)(2012)}}{-6}$. Since we want $x$ being positive, we use $\negative$ (Error compiling LaTeX. Unknown error_msg) for $\pm$ to give us $x>\frac{-9-\sqrt{24225}}{-6}$. The RHS is the same as $\frac{9+\sqrt{24225}}{6}$. Our goal now is to find $\sqrt{24225}$ approximately. Note that $155^2=\underline{15*16},2,5=24025$ (where $15*16, 2$, and $5$ are digits of $155^2$), therefore $156>\sqrt{24225}>155$, and substitute this into our equation for $x$ to give a smallest possible value of $x$ being $\frac{156+9}{6}>x>\frac{155+9}{6}$ to give us $27.5>x>27.\overline{3}$ and hence the smallest possible positive integer value for $x$ is $\boxed{028}$.

Solution 2

Another way of finding $f(x)$ is noticing that $a_n-a_{n-1}=\sum_{i=0}^{n}f(i)-\sum_{i=0}^{n-1}f(i)= f(n)$ and that the terms of the arithmetic sequence are $a_n+(n-1)^2$. The difference of arithmetic sequences are constant, so $a_n+(n-1)^2-(a_{n-1}+(n-2)^2)= f(n) + 3n^2 - 9n + 7$ is constant. Let that constant be $c$. Then $f(n)+3n^2-9n+7 =c$ and so $f(n) = -3n^2+9n +7+c$. The solution follows as above.

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