Mock AIME 2 Pre 2005 Problems/Problem 9

Solution

We begin by determining the value of $k_{1997}$ . Experimenting, we find the first few $k_{i}$ s: $k_{1} = 3, k_{2} = {3^2}, k_{3} = {3^2}+3, k_{4} = {3^3}...$

We observe that because ${3^k}>\sum_{n=1}^{k-1} {3^n}$ , $k_{i}$ will be determined by the base 2 expansion of i. Specifically, every 1 in the $2^{(n-1)}$ s digit of the expansion corresponds to adding $3^n$ to $k_{i}$ . Since $1997 = 11111001101$ base 2, $k_{1997}={3^11}+{3^10}+{3^9}+{3^8}+{3^7}+{3^4}+{3^3}+{3^1}.$

Now we look for ways to attain an element with degree $k_{1997}$ . Since each sum of powers of 3 is unique, there is only one; namely, take the x element for every binomial with a degree of one of the added powers of 3 in $k_{1997}$ , and the 1 for all else. Finally, since the coefficients of the x elements are equal to the degree to which the 3 is raised, we conclude $a_{1997}=11*10*9*8*7*4*3*1 =665280 =>\boxed{280}$

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Mock AIME 2 Pre 2005 Problems/Problem 9

Solution