Difference between revisions of "2014 AMC 10A Problems/Problem 24"
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<math>13,14,15,16,17,18</math> etc. | <math>13,14,15,16,17,18</math> etc. | ||
| − | 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 = 499494</math>) The last number of the <math>996</math>th row (with 999 terms) is <math>499494 + (1+2+3+4.....+996)= 996000</math>, (we add the <math>1-996</math> because of | + | 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 = 499494</math>) The last number of the <math>996</math>th row (with 999 terms) is <math>499494 + (1+2+3+4.....+996)= 996000</math>, (we add the <math>1-996</math> because of the numbers we skip) so our answer is <math>996000 + 506 = \boxed{(A)996506}</math> |
==See Also== | ==See Also== | ||
Revision as of 22:39, 6 February 2014
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
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 (with 999 terms) is 
, (we add the 
because of the numbers we skip) so our 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. 
