The UKMT Senior Challenge-II

Number Theory Level 4

The increasing sequence 1, 3, 4, 9, 10, 12, 13, … contains all the powers of 3 and all the numbers that can be written as the sum of two or more distinct powers of 3. What is the 70th number in the sequence?

This problem is not original.This problem is part of this set .


The answer is 741.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Chew-Seong Cheong
Mar 11, 2015

The n t h n^{th} term a n a_n is given by converting n n into base 2 2 and then replacing the powers of 2 2 with powers of 3 3 . See below:

n = 1 = 2 0 ⇒ a 1 = 3 0 = 1 n = 2 = 2 1 ⇒ a 2 = 3 1 = 3 n = 3 = 2 1 + 2 0 ⇒ a 3 = 3 1 + 3 0 = 4 n = 4 = 2 2 ⇒ a 4 = 3 2 = 9 n = 5 = 2 2 + 2 0 ⇒ a 5 = 3 2 + 3 0 = 10 n = 6 = 2 2 + 2 1 ⇒ a 6 = 3 2 + 3 1 = 12 n = 7 = 2 2 + 2 1 + 2 0 ⇒ a 3 = 3 2 + 3 1 + 3 0 = 13 . . . ⇒ n = 70 = 2 6 + 2 2 + 2 1 ⇒ a 3 = 3 6 + 3 2 + 3 1 = 741 \begin{array}{llll} & n = 1 = 2^0 & \Rightarrow a_1 = 3^0 & = 1 \\ & n = 2 = 2^1 & \Rightarrow a_2 = 3^1 & = 3 \\ & n = 3 = 2^1+2^0 & \Rightarrow a_3 = 3^1+3^0 & = 4 \\ & n = 4 = 2^2 & \Rightarrow a_4 = 3^2 & = 9 \\ & n = 5 = 2^2+2^0 & \Rightarrow a_5 = 3^2+3^0 & = 10 \\ & n = 6 = 2^2+2^1 & \Rightarrow a_6 = 3^2+3^1 & = 12 \\ & n = 7 = 2^2+2^1+2^0 & \Rightarrow a_3 = 3^2+3^1+3^0 & = 13 \\ & & \quad ... \\ \Rightarrow & n = 70 = 2^6+2^2+2^1 & \Rightarrow a_3 = 3^6+ 3^2+3^1 & = \boxed{741} \end{array}

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That's exactly how I did it because I noticed that if the power is n {n} then it will be used 2 n 2^{n} times.

Curtis Clement - 6 years, 3 months ago

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