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?

$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$

Solution 1

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 because $4+5+6+7......+999 = 499,494$ . The last number of the $996$ th row (when including the numbers skipped) is $499,494 + (1+2+3+4.....+996)= 996,000$ , (we add the $1-996$ because of the numbers we skip) so our answer is $996,000 + 506 = \boxed{\textbf{(A)}996,506}$ .

Solution 2

Let's start with natural numbers, with no skips in between.

$1,2,3,4,5,...,500,000$

All we need to do is count how many numbers are skipped, $n$ , and "push" (add on to) $500,000$ however many numbers are skipped.

Clearly, $\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}$ . This means that the number of skipped number "blocks" in the sequence is $999-3=996$ because we started counting from 4.

Therefore $n=\frac{996(997)}{2}=496,506$ , and the answer is $496,506+500000=\boxed{\textbf{(A)}996,506}$ .

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2014 AMC 10A Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

See Also