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 (with 999 terms) is $499494 + (1+2+3+4.....+996)= 996000$ , (we add the $1~996$ because of our rests) so our answer is $996000 + 506 = \boxed{(A)996506}$

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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2014 AMC 10A Problems/Problem 24

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