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

## 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 $996000 + 506 = \boxed{\textbf{(A)}996506}$