2014 AMC 10A Problem 24

Track two running totals after $n$ complete rounds.

Numbers listed: $\sum_{k=1}^{n} (k + 3) = \frac{n(n+1)}{2} + 3n = \frac{n(n+7)}{2}$.

Integers passed (listed or skipped): $\sum_{k=1}^{n} (2k + 3) = n(n+1) + 3n = n(n+4)$. So round $n+1$ begins with the integer $n(n+4) + 1$.

Check: $n = 1$ gives $4$ listed and $5$ passed; $n = 2$ gives $9$ listed and $12$ passed, and the next term is $13$, matching the given start.

Now find the round containing the $500{,}000$th term. We need $\frac{n(n+7)}{2} \approx 500{,}000$, i.e. $n^2 + 7n \approx 1{,}000{,}000$, so $n \approx 996$. Indeed
$$\frac{996 \cdot 1003}{2} = 498 \cdot 1003 = 499{,}494 < 500{,}000,$$
and round $997$ lists $1000$ more, which is plenty. The integers passed after $996$ rounds total $996 \cdot 1000 = 996{,}000$.

Round $997$ lists $1000$ consecutive integers starting at $996{,}001$. The $500{,}000$th term is entry number $500{,}000 - 499{,}494 = 506$ of this block:
$$996{,}000 + 506 = 996{,}506.$$

The answer is $\boxed{\textbf{(A)}\ 996,506}$.

Block-structured sequences call for two cumulative counts: how many terms have been produced and how far along the integers we have moved. Both are quadratic in the round number, so a square-root estimate locates the right round and exact arithmetic pins down the position inside it. The nearly identical answer choices are a warning that the final off-by-one step is where the problem is actually decided; verifying the formulas on the first two rounds guards against it.

Complementary view: the $500{,}000$th listed number equals $500{,}000$ plus the number of skipped integers before it. After $996$ rounds, $\frac{996 \cdot 997}{2} = 496{,}506$ integers were skipped, and no more are skipped until round $997$ finishes. So the answer is $500{,}000 + 496{,}506 = 996{,}506$.

• Off-by-one at the start of round 997: the block begins at 996,001, so its 506th entry is 996,506, not 996,505 or 996,507.
• Confusing the two totals, for example solving $n(n+4) = 500{,}000$ to choose the round, or forgetting the skipped numbers when converting position to value.
• Arithmetic slips in $498 \cdot 1003$; note $996 \cdot 1000$ is exact, which makes the passed-count especially clean.