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 = | + | 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> |
==See Also== | ==See Also== |
Revision as of 23:49, 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 (when including the numbers skipped) 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 |
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