Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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Therefore <math>n=\frac{996(997)}{2}=496,506</math>, and the answer is <math>496,506+500000=\boxed{\textbf{(A)}996,506}</math>. | Therefore <math>n=\frac{996(997)}{2}=496,506</math>, and the answer is <math>496,506+500000=\boxed{\textbf{(A)}996,506}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s | ||
==See Also== | ==See Also== |
Revision as of 08:27, 14 April 2020
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 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 because . 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 .
Video Solution
https://www.youtube.com/watch?v=KfGtE4G6tBo&t=427s
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.