Difference between revisions of "2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"
(→Solution) |
(→Solution) |
||
Line 6: | Line 6: | ||
== 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 \boxed<math>1024</math>. |
== See also == | == See also == |
Revision as of 00:17, 1 February 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 \boxed
.
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 |