Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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If we list the rows by iterations, then we get | If we list the rows by iterations, then we get | ||
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so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row. (<math>4+5+6+7......+999 = 499,494</math>) The last number of the <math>996</math>th row (when including the numbers skipped) is <math>499,494 + (1+2+3+4.....+996)= 996,000</math>, (we add the <math>1-996</math> because of the numbers we skip) so our answer is <math>996,000 + 506 = \boxed{\textbf{(A)}996,506}</math> | so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row. (<math>4+5+6+7......+999 = 499,494</math>) The last number of the <math>996</math>th row (when including the numbers skipped) is <math>499,494 + (1+2+3+4.....+996)= 996,000</math>, (we add the <math>1-996</math> because of the numbers we skip) so our answer is <math>996,000 + 506 = \boxed{\textbf{(A)}996,506}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Let's start with natural numbers, with no skips in between. | ||
| + | |||
| + | <math>1,2,3,4,5,...,500000</math> | ||
| + | |||
| + | All we need to do is count how many numbers are skipped, <math>n</math>, and "push" (add on) <math>500000</math> however many numbers are skipped. | ||
| + | |||
| + | Clearly, <math>\frac{999(1000)}{2}\le500,000</math>. This means that the number skipped number "blocks" in the sequence is <math>999-3=996</math> because we started counting from 4. | ||
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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>. | ||
==See Also== | ==See Also== | ||
Revision as of 17:12, 8 February 2014
Contents
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) 
however many numbers are skipped.
Clearly, 
. This means that the number 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. 
