Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | A sequence of natural numbers is constructed by listing the first <math>4</math>, then skipping one, listing the next <math>5</math>, skipping <math>2</math>, listing <math>6</math>, skipping <math>3</math>, and, on the <math>n</math>th iteration, listing <math>n+3</math> and skipping <math>n</math>. The sequence begins <math>1,2,3,4,6,7,8,9,10,13</math>. What is the <math>500,\!000</math>th number in the sequence? | + | A sequence of natural numbers is constructed by listing the first <math>4</math>, then skipping one, listing the next <math>5</math>, skipping <math>2</math>, listing <math>6</math>, skipping <math>3</math>, and, on the <math>n</math>th iteration, listing <math>n+3</math> and skipping <math>n</math>. The sequence begins <math>1,2,3,4,6,7,8,9,10,13</math>. What is the <math>500,\!000</math>th the number in the sequence? |
<math> \textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510 </math> | <math> \textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510 </math> |
Revision as of 09:52, 30 January 2016
Contents
[hide]Problem
A sequence of natural numbers is constructed by listing the first , then skipping one, listing the next , skipping , listing , skipping , and, on the th iteration, listing and skipping . The sequence begins . What is the th the number in the sequence?
Solution 1
If we list the rows by iterations, then we get
etc.
so that the th number is the th number on the th row. () The last number of the th row (when including the numbers skipped) is , (we add the because of the numbers we skip) so our answer is
Solution 2
Let's start with natural numbers, with no skips in between.
All we need to do is count how many numbers are skipped, , and "push" (add on to) however many numbers are skipped.
Clearly, . This means that the number of skipped number "blocks" in the sequence is because we started counting from 4.
Therefore , and the answer is .
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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