# Mock AIME II 2012 Problems/Problem 15

## Problem:

Define for and . Given that is a polynomial, and is an arithmetic sequence, find the smallest positive integer value of such that .

## Solution 1

**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. ! Undefined control sequence.) 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 .

## Solution 2

Another way of finding is noticing that and that the terms of the arithmetic sequence are . The difference of arithmetic sequences are constant, so is constant. Let that constant be . Then and so . The solution follows as above.