Difference between revisions of "2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"

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== Solution==
 
== Solution==
 
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Since there is <math>1</math> one, <math>2</math> twos, <math>4</math> fours and so on, the number that occupies a position is <math>1+2+4+8+ \cdot \cdot \cdot</math>. Since <math>2^0+2^1+2^2+2^3+ \cdot \cdot \cdot +2^{n-1} = 2^n-1</math>, we get the <math>1023</math>rd number to be <math>512</math>. Since the <math>2015</math>th position is <math>992</math> positions higher than the last <math>512</math>, the answer is <math>\boxed{1024}</math>.
 
 
  
 
== See also ==
 
== See also ==
 
{{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}
 
{{UNM-PNM Math Contest box|year=2015|n=II|before=First question|num-a=2}}
  
[[Category:]]
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[[Category:Beginning Algebra Problems]]

Latest revision as of 23:18, 31 January 2020

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
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