SAT1000 - P904

Algebra Level pending

If we define S = { a l 1 , a l 2 , , a l n } S=\{a_{l_1},a_{l_2}, \cdots ,a_{l_n}\} as the k k th subset for set E = { a 1 , a 2 , , a 10 } E=\{a_1,a_2,\cdots,a_{10}\} , where k = i = 1 n 2 l i 1 k=\displaystyle \sum_{i=1}^{n} 2^{l_{i}-1} .

Then what is the 211 211 th subset for E E ?

How to submit:

Sort the indices of the elements of S S from smallest to largest and put them together. For example, if S = { a 1 , a 2 , a 3 } S=\{a_1,a_2,a_3\} , then submit 123 123 , and when S = { a 1 , a 2 , a 10 } S=\{a_1,a_2,a_{10}\} , then submit 1210 1210 .


Have a look at my problem set: SAT 1000 problems


The answer is 12578.

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1 solution

David Vreken
Jul 1, 2020

211 = 2 0 + 2 1 + 2 4 + 2 6 + 2 7 = 2 1 1 + 2 2 1 + 2 5 1 + 2 7 1 + 2 8 1 211 = 2^{0} + 2^{1} + 2^{4} + 2^{6} + 2^{7} = 2^{1 - 1} + 2^{2 - 1} + 2^{5 - 1} + 2^{7 - 1} + 2^{8 - 1} , so S = { a 1 , a 2 , a 5 , a 7 , a 8 } S = \{a_1, a_2, a_5, a_7, a_8\} , which means the answer to submit is 12578 \boxed{12578} .

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