Difference between revisions of "2014 AMC 10A Problems/Problem 24"

(Solution)
(Solution)
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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 (when including the numbers skipped) 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>
+
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 (when including the numbers skipped) 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{\textbf{(A)}996506}</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 $4$, then skipping one, listing the next $5$, skipping $2$, listing $6$, skipping $3$, and, on the $n$th iteration, listing $n+3$ and skipping $n$. The sequence begins $1,2,3,4,6,7,8,9,10,13$. What is the $500,000$th number in the sequence?

$\textbf{(A)}\ 996,506\qquad\textbf{(B)}\ 996507\qquad\textbf{(C)}\ 996508\qquad\textbf{(D)}\ 996509\qquad\textbf{(E)}\ 996510$


Solution

If we list the rows by iterations, then we get

$1,2,3,4$

$6,7,8,9,10$

$13,14,15,16,17,18$ etc.

so that the $500,000$th number is the $506$th number on the $997$th row. ($4+5+6+7......+999 = 499494$) The last number of the $996$th row (when including the numbers skipped) is $499494 + (1+2+3+4.....+996)= 996000$, (we add the $1-996$ because of the numbers we skip) so our answer is $996000 + 506 = \boxed{\textbf{(A)}996506}$

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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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