The increasing sequence $1; 3; 4; 9; 10; 12; 13; 27; 28; 30; 31, \ldots$ is formed with positive integers which are powers of $3$ or sums of different powers of $3$. Which number is in the $100^{th}$ position?
Problem
Source:
Tags: MATHCOUNTS, number theory proposed, number theory
mahanmath
29.04.2012 20:00
Write in base $2$ and read in base $3$ , that's all !
ssilwa
21.12.2013 18:51
Here is a quote from a MathCounts Discussion on Puerto Rico. tonypr wrote: Something to keep in mind: While the cut off is usually lower than average, the test is only given in English. Consider the fact that roughly 15% of Puerto Rico's population speaks and understands English well. This severely limits the number of people that can do well on the test. (Consider having to take an entire math test in Spanish, given that English is your first language.)
lucasxia01
27.06.2016 01:51
batter8642 wrote: The increasing sequence $1; 3; 4; 9; 10; 12; 13; 27; 28; 30; 31, \ldots$ is formed with positive integers which are powers of $3$ or sums of different powers of $3$. Which number is in the $100^{th}$ position?
Let's write this in base 3. Our sequence now becomes:
$1, 10, 11, 100, 101, 110, 111, 1000, \dots$. In base 2, it becomes even nicer:
$1, 2, 3, 4, 5, 6, 7, 8, \dots$. We see that 100 is in the 100th position which is 1100100 in the first sequence. This is in base 3 so we convert back to base 10 to get $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed{981}$.
toluene
27.06.2016 05:07
It's No.7 problem of AIME in 1986:)
ike.chen
02.07.2021 06:34
(Storage) Since each power of $3$ can only appear up to once within each term, we know the sequence contains all positive integers in base $3$ with digits $0$ or $1$. This reminds us of binary numbers. Now, it easily follows that the hundredth term has the same digits (not value) as the binary representation of $100$, or $1100100_{2}$, but in base $3$. Thus, $$a_{100} = 1100100_{3} = 3^6 + 3^5 + 3^2 = \boxed{981}.$$