Difference between revisions of "2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"
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== Solution== | == Solution== | ||
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>1024</math>. | 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>1024</math>. | ||
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== See also == | == See also == |
Revision as of 17:58, 4 January 2020
Problem
In the sequence what number occupies position ?
Solution
Since there is one, twos, fours and so on, the number that occupies a position is . Since , we get the rd number to be . Since the th position is positions higher than the last , the answer is .
See also
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 |