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 $\boxed{1024}$ .

2015 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions

Page

Toolbox

Search

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

Problem

Solution

See also