Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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All we need to do is count how many numbers are skipped, <math>n</math>, and "push" (add on to) <math>500000</math> however many numbers are skipped. | All we need to do is count how many numbers are skipped, <math>n</math>, and "push" (add on to) <math>500000</math> however many numbers are skipped. | ||
− | Clearly, <math>\frac{999(1000)}{2}<500,000</math>. This means that the number of skipped number "blocks" in the sequence is <math>999-3=996</math> because we started counting from 4. | + | Clearly, <math>\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}</math>. This means that the number of skipped number "blocks" in the sequence is <math>999-3=996</math> because we started counting from 4. |
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>. |
Revision as of 16:46, 26 December 2015
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.