# Difference between revisions of "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 (Problems • Answer Key • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNM-PNM Problems and Solutions
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