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[[Category:Intermediate Combinatorics Problems]]

Revision as of 09:55, 20 May 2015

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. ($4+5+6+7......+999 = 499,494$) 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.

$1,2,3,4,5,...,500000$

All we need to do is count how many numbers are skipped, $n$, and "push" (add on to) $500000$ however many numbers are skipped.

Clearly, $\frac{999(1000)}{2}\le500,000$. 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 $496,506+500000=\boxed{\textbf{(A)}996,506}$.

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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