2013/14 AHSMC Part 1 Problem 16

Problem 16 of 16

The list $1$ , $3$ , $4$ , $9$ , $10$ , $12$ , $13$ , … contains in increasing order all positive integers which can be expressed as sums of one or more distinct integral powers of $3$ . The $100$ -th number in this list is

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Solution

If we switch the powers of $3$ to the powers of $2$ , then we get all the positive integers. Hence we convert $100$ into base $2$ , obtaining $100 = 2^6 + 2^5 + 2^2$ . It follows that the $100$ -th number on the list is $3^6 + 3^5 + 3^2 = 981$ . The answer is (A).

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