2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1

Problem

In the sequence $1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, \cdots$ what number occupies position $2015$?

Solution

Since there is $1$ one, $2$ twos, $4$ fours and so on, the number that occupies a position is $1+2+4+8+ \cdot \cdot \cdot$. Since $2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1$, we get the $1023$rd number to be $512$. Since the $2015$th position is $992$ positions higher than the last $512$, the answer is $\boxed{1024}$.

See also

2015 UNM-PNM Contest II (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNM-PNM Problems and Solutions
Invalid username
Login to AoPS