AMC 10 · 2014 · #24
Grade 6 arithmeticA 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 nth iteration, listing n+3 and skipping n. The sequence begins 1,2,3,4,6,7,8,9,10,13. What is the 500,000th number in the sequence?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A list of counting numbers is built in rounds. In round $n$ you write down the next $n+3$ numbers, then jump over (skip) the next $n$ numbers. So round 1 writes $4$ numbers and skips $1$, round 2 writes $5$ and skips $2$, and so on, giving $1,2,3,4,\;6,7,8,9,10,\;13,\dots$ Find the $500{,}000$th number that actually gets written down.
Givens: Round $n$ writes $n+3$ consecutive natural numbers, then skips the next $n$ numbers; The written list starts $1,2,3,4,6,7,8,9,10,13,\dots$; We want the value of the $500{,}000$th written number; Answer choices: (A) $996{,}506$, (B) $996{,}507$, (C) $996{,}508$, (D) $996{,}509$, (E) $996{,}510$
Unknowns: The actual value of the $500{,}000$th number in the written list
Understand
Restated: A list of counting numbers is built in rounds. In round $n$ you write down the next $n+3$ numbers, then jump over (skip) the next $n$ numbers. So round 1 writes $4$ numbers and skips $1$, round 2 writes $5$ and skips $2$, and so on, giving $1,2,3,4,\;6,7,8,9,10,\;13,\dots$ Find the $500{,}000$th number that actually gets written down.
Givens: Round $n$ writes $n+3$ consecutive natural numbers, then skips the next $n$ numbers; The written list starts $1,2,3,4,6,7,8,9,10,13,\dots$; We want the value of the $500{,}000$th written number; Answer choices: (A) $996{,}506$, (B) $996{,}507$, (C) $996{,}508$, (D) $996{,}509$, (E) $996{,}510$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #4 Introduce a Variable, #6 Guess and Check, #16 Change Focus / Count the Complement
Tool #5 (Look for a Pattern) turns the wordy rule into a clean block structure: round $n$ contributes $n+3$ written numbers and $n$ skipped numbers. Tool #4 (Introduce a Variable) names the round number $k$ and builds a formula for how many numbers have been written after $k$ rounds, so we do not have to list a million terms. Tool #6 (Guess and Check) plugs $k=996$ and $k=997$ into that formula to pin down which round holds the $500{,}000$th written number. Finally Tool #16 (Change Focus) flips the question to the skipped numbers: the value of a written number equals its position plus how many numbers were skipped before it, and that skipped total is a short triangular-number computation.
Execute — Answer: A
4.OA.C.5 Step 1 Read the round pattern
- Group the sequence into rounds.
- Round $n$ writes $n+3$ numbers, then skips $n$.
- So the written blocks have sizes $4,5,6,7,\dots$ and the gaps between blocks have sizes $1,2,3,\dots$ Writing then skipping, round $n$ uses up $(n+3)+n = 2n+3$ natural numbers in total before the next round starts.
💡 Naming one repeating round lets you describe the whole endless list with a single rule.
6.EE.B.6 Step 2 Count written numbers after k rounds
- Let $k$ be a round number.
- The total written after $k$ full rounds is $4+5+6+\dots+(k+3)$.
- That is the sum $\sum_{n=1}^{k}(n+3)=\frac{k(k+1)}{2}+3k=\frac{k^2+7k}{2}$.
- Call this $L(k)$: the count of written numbers once round $k$ is finished.
💡 A running total of block sizes is just the triangular-number sum with a constant added to each term.
6.EE.A.2 Step 3 Find the round holding the 500,000th number
- We need $L(k)$ to reach $500{,}000$, so solve $\frac{k^2+7k}{2}\approx 500{,}000$, i.e.
- $k^2+7k\approx 1{,}000{,}000$, giving $k\approx 996.5$.
- Test the two nearby whole rounds.
- $L(996)=\frac{996^2+7\cdot996}{2}=\frac{992016+6972}{2}=499{,}494$.
- $L(997)=\frac{997^2+7\cdot997}{2}=\frac{994009+6979}{2}=500{,}494$.
- Since $499{,}494 < 500{,}000 \le 500{,}494$, the $500{,}000$th written number lands inside round $997$'s written block.
💡 Evaluating the total at two consecutive rounds brackets the target between them.
5.NBT.B.5 Step 4 Switch focus to the skipped numbers
- Here is the key flip: every natural number up to the $500{,}000$th written one is either written or skipped, so its value equals its position ($500{,}000$) plus the count of numbers skipped before it.
- The $500{,}000$th written number sits inside round $997$'s block, which comes before round $997$ does any skipping.
- So the only skips that happened first are from rounds $1$ through $996$: that total is $S=1+2+\dots+996=\frac{996\cdot997}{2}=496{,}506$.
💡 A written number's value is just how far into the natural numbers you have walked: steps taken written, plus steps jumped over.
4.NBT.B.4 Step 5 Combine position and skips
- Add the skipped count back onto the position: value $=500{,}000+496{,}506=996{,}506$.
- That matches choice (A).
- Quick sanity note on the units digit: the skipped total ends in $6$, and $500{,}000$ ends in $0$, so the answer must end in $6$ — only (A) does.
- The $500{,}000$th number in the sequence is $\textbf{(A)}\ 996{,}506$.
💡 Position plus everything skipped before it gives the true value with one addition.
4.OA.C.5 Group the sequence into rounds. Round $n$ writes $n+3$ numbers, then skips $n$. 6.EE.B.6 Let $k$ be a round number. The total written after $k$ full rounds is $4+5+6+\do 6.EE.A.2 We need $L(k)$ to reach $500{,}000$, so solve $\frac{k^2+7k}{2}\approx 500{,}000 5.NBT.B.5 Here is the key flip: every natural number up to the $500{,}000$th written one i 4.NBT.B.4 Add the skipped count back onto the position: value $=500{,}000+496{,}506=996{,} Review
Reasonableness: Cross-check by tracking the natural numbers directly. After round $996$, the numbers used up (written plus skipped) total $L(996)+S=499{,}494+496{,}506=996{,}000$, so round $997$ starts writing at $996{,}001$. Round $997$ writes $1000$ numbers, taking written-positions $499{,}495$ through $500{,}494$. The $500{,}000$th written number is the $(500{,}000-499{,}494)=506$th of these, with value $996{,}001+505=996{,}506$. Both routes agree, and the answer sits right in the middle of the choice cluster, exactly where a value near $996{,}500$ should land. So (A) is confirmed.
Alternative: Skip the round-by-round bookkeeping and use one estimate: the value of the $m$th written number is about $m$ plus roughly $m/2$ skipped numbers (each round skips about half as many as it writes), so around $1.5\times500{,}000=750{,}000$ as a rough scale, then tighten with the exact triangular sums. More cleanly, note value $=$ position $+$ skips, write skips as a triangular number $\frac{k(k+1)}{2}$, and solve for the $k$ that places position $500{,}000$ inside round $k+1$; that gives $k=996$ and the same $996{,}506$.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Turning the written-then-skip rule into a repeating round structure of block sizes $n+3$ and gaps $n$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Introducing the round variable $k$ and deriving $L(k)=\frac{k^2+7k}{2}$ for the written count.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $L(996)=499{,}494$ and $L(997)=500{,}494$ to locate the round containing the target.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing the skipped total $\frac{996\cdot997}{2}=496{,}506$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding position and skips, $500{,}000+496{,}506=996{,}506$, and the cross-check $499{,}494+496{,}506=996{,}000$.)
⭐ A written number's value is where you are in the counting numbers: add up everything you skipped and stick it onto your position.
⭐ A written number's value is where you are in the counting numbers: add up everything you skipped and stick it onto your position.
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