Difference between revisions of "2014 AMC 10A Problems/Problem 24"
Fridaychimp (talk | contribs) (→Solution 1) |
Alexwin0806 (talk | contribs) m (→Solution 1) |
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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 | + | so that the <math>500,000</math>th number is the <math>506</math>th number on the <math>997</math>th row because <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== | ==Solution 2== |
Revision as of 18:41, 9 March 2020
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 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 .
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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