Mock AIME II 2012 Problems/Problem 15


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. ) 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 $a_n+(n-1)^3$. The difference of arithmetic sequences are constant, so $a_n+(n-1)^3-(a_{n-1}+(n-2)^3)= f(n) + 3n^2 - 9n + 7$ 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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