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Difference between revisions of "Mock AIME 2 Pre 2005 Problems/Problem 9"

(Created page with "== Solution == We begin by determining the value of <math>k_{1997}</math>. Experimenting, we find the first few <math>k_{i}</math>s: <cmath>k_{1} = 3, k_{2} = {3^2}, k_{3} =...")
 
(→‎Solution)
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We observe that because <math>{3^k}>\sum_{n=1}^{k-1} {3^n}</math>, <math>k_{i}</math> will be determined by the base 2 expansion of i. Specifically, every 1 in the <math>2^{(n-1)}</math>s digit of the expansion corresponds to adding <math>3^n</math> to <math>k_{i}</math>. Since <math>1997 = 11111001101</math> base 2,  
 
We observe that because <math>{3^k}>\sum_{n=1}^{k-1} {3^n}</math>, <math>k_{i}</math> will be determined by the base 2 expansion of i. Specifically, every 1 in the <math>2^{(n-1)}</math>s digit of the expansion corresponds to adding <math>3^n</math> to <math>k_{i}</math>. Since <math>1997 = 11111001101</math> base 2,  
−
<cmath>k_{1997}={3^11}+{3^10}+{3^9}+{3^8}+{3^7}+{3^4}+{3^3}+{3^1}.</cmath>
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<cmath>k_{1997}={3^{11}}+{3^{10}}+{3^9}+{3^8}+{3^7}+{3^4}+{3^3}+{3^1}.</cmath>
  
 
Now we look for ways to attain an element with degree <math>k_{1997}</math>. 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 <math>k_{1997}</math>, 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
 
Now we look for ways to attain an element with degree <math>k_{1997}</math>. 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 <math>k_{1997}</math>, 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

Revision as of 15:00, 31 December 2020

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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