AMC 10 · 2020 · #11
Grade 6 arithmeticProblem
What is the median of the following list of 4040 numbers?
1,2,3,…,2020,12,22,32,…,20202
Pick an answer.
AMC 10 2020 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: Sort the $4040$-number list $\{1, 2, 3, \ldots, 2020,\; 1^2, 2^2, 3^2, \ldots, 2020^2\}$ in non-decreasing order, then find the median (the average of the $2020$-th and $2021$-st terms).
Givens: The first $2020$ entries are the integers $1, 2, \ldots, 2020$; The next $2020$ entries are the squares $1^2, 2^2, \ldots, 2020^2$; Total count $= 4040$, so the median is the average of the $2020$-th and $2021$-st sorted values; Answer choices: $1974.5,\; 1975.5,\; 1976.5,\; 1977.5,\; 1978.5$
Unknowns: Median of the combined sorted list
Understand
Restated: Sort the $4040$-number list $\{1, 2, 3, \ldots, 2020,\; 1^2, 2^2, 3^2, \ldots, 2020^2\}$ in non-decreasing order, then find the median (the average of the $2020$-th and $2021$-st terms).
Givens: The first $2020$ entries are the integers $1, 2, \ldots, 2020$; The next $2020$ entries are the squares $1^2, 2^2, \ldots, 2020^2$; Total count $= 4040$, so the median is the average of the $2020$-th and $2021$-st sorted values; Answer choices: $1974.5,\; 1975.5,\; 1976.5,\; 1977.5,\; 1978.5$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #2 Make a Systematic List, #5 Look for a Pattern, #3 Eliminate Possibilities
The list is huge ($4040$ entries), so Tool #9 (Easier Problem): first solve the same setup with $N=4$ in place of $2020$ and watch what happens. From the small case we see that for each integer $k$, the squares $1^2, 2^2, \ldots, k_{\max}^2$ that are $\le k$ slip in among the small numbers and "push" the median value down. Tool #2 (Systematic List): count exactly how many list entries are $\le k$ for each candidate $k$ near the middle. Tool #5 (Pattern): the rank of $k$ equals $k + (\text{number of squares} \le k)$. Tool #3 (Eliminate): match the final median to the five choices.
Execute — Answer: C
6.SP.A.3 Step 1 - Try a tiny version first: $N=4$.
- The list is $\{1, 2, 3, 4, 1, 4, 9, 16\}$, which sorted is $1, 1, 2, 3, 4, 4, 9, 16$.
- The $4$-th and $5$-th terms are $3$ and $4$, so the median is $3.5$.
- Notice the median sits among the small integers, not among the big squares.
💡 Grade 6 median: shrink the problem to a list you can sort by hand.
5.NBT.B.5 Step 2 - Generalize the lesson from the small case.
- To find the rank of a value $k$ in the sorted list, count how many entries are $\le k$: it's $k$ from the integers plus the number of squares $1^2, 2^2, \ldots$ that are $\le k$.
- For our problem with $N = 2020$, the squares $\le 2020$ are exactly $1^2, 2^2, \ldots, 44^2$ (since $44^2 = 1936 \le 2020 < 2025 = 45^2$).
💡 Grade 5 multi-digit multiplication: $44 \times 44$ and $45 \times 45$ pin down the squares that fit.
4.OA.C.5 Step 3 - Build the counting pattern.
- For any integer $k$ between $44^2 = 1936$ and $45^2 = 2025$, exactly $44$ squares are $\le k$.
- So the rank of $k$ in the sorted list is $k + 44$.
💡 Grade 4 pattern rule: each integer in that band picks up the same $44$ extra entries from squares.
4.OA.A.3 Step 4 - Find the $2020$-th term.
- Solve $k + 44 = 2020$, so $k = 1976$.
- Check: $1976$ is in our integer list, and there are exactly $1976$ integers $\le 1976$ plus $44$ squares $\le 1976$, total $2020$.
- So the $2020$-th sorted entry is $1976$.
💡 Grade 4 word-problem subtraction: $2020 - 44 = 1976$ undoes the rank shift.
4.NBT.A.2 Step 5 - Find the $2021$-st term.
- The next candidate is the next integer $1977$ (still less than the next square $45^2 = 2025$), so $S_{2021} = 1977$.
💡 Grade 4 comparing whole numbers: $1977 < 2025$, so $1977$ comes first.
6.SP.A.3 Step 6 - Average the two middle terms to get the median.
- $(1976 + 1977) / 2 = 3953/2 = 1976.5$.
💡 Grade 6 median: average the two middle entries of an even-length sorted list.
4.NBT.A.2 Step 7 - Match to the choices: $1976.5$ is option (C).
- The other choices ($1974.5, 1975.5, 1977.5, 1978.5$) would each require an off-by-one or off-by-two miscount of squares.
💡 Grade 4 comparison: only one choice equals $1976.5$.
6.SP.A.3 Try a tiny version first: $N=4$. The list is $\{1, 2, 3, 4, 1, 4, 9, 16\}$, whic 5.NBT.B.5 Generalize the lesson from the small case. To find the rank of a value $k$ in th 4.OA.C.5 Build the counting pattern. For any integer $k$ between $44^2 = 1936$ and $45^2 4.OA.A.3 Find the $2020$-th term. Solve $k + 44 = 2020$, so $k = 1976$. Check: $1976$ is 4.NBT.A.2 Find the $2021$-st term. The next candidate is the next integer $1977$ (still le 6.SP.A.3 Average the two middle terms to get the median. $(1976 + 1977) / 2 = 3953/2 = 19 4.NBT.A.2 Match to the choices: $1976.5$ is option (C). The other choices ($1974.5, 1975.5 Review
Reasonableness: All five answer choices cluster around $1976.5$, hinting the median should be near the integer whose rank shifts by about $44$ (the count of squares $\le 2020$). Our value $1976.5$ sits exactly $43.5$ below $2020$, matching the $44$-square shift up to the half-step from averaging two consecutive integers. Magnitude and sign are both reasonable.
Alternative: Tool #16 (Complement / Change Focus): instead of counting from the bottom, count how many list entries are $> 2020$. The big squares $45^2, 46^2, \ldots, 2020^2$ give $2020 - 44 = 1976$ entries above $2020$. So out of $4040$ entries, $1976$ are above $2020$ and $2064$ are at most $2020$. The two middle ranks $2020, 2021$ both fall in the at-most-$2020$ block, $44$ entries from its top — landing on $1976, 1977$.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing $1977 < 2025$ to identify the next sorted entry and matching the final number to the choices.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Subtracting $2020 - 44 = 1976$ to recover the integer whose rank is $2020$.)4.OA.C.5Generate a number or shape pattern following a given rule (Stating the pattern $\text{rank}(k) = k + 44$ for integers $k$ in $[1936, 2024]$.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Computing $44 \times 44 = 1936$ and $45 \times 45 = 2025$ to bracket the squares $\le 2020$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Defining the median of an even-length list as the average of the two middle entries.)
⭐ This AMC 10 problem only needs Grade 6 median you already know! The trick: the squares $1, 4, 9, 16, \ldots$ that stay below $2020$ are exactly $1^2$ through $44^2$ (since $44^2 = 1936$ but $45^2 = 2025$ is too big). So between $1936$ and $2024$, every integer $k$ has rank $k + 44$. Solving $k + 44 = 2020$ gives $k = 1976$, the $2020$-th entry. The $2021$-st is $1977$. Median $= (1976 + 1977)/2 = \mathbf{1976.5}$, answer (C).
⭐ This AMC 10 problem only needs Grade 6 median you already know! The trick: the squares $1, 4, 9, 16, \ldots$ that stay below $2020$ are exactly $1^2$ through $44^2$ (since $44^2 = 1936$ but $45^2 = 2025$ is too big). So between $1936$ and $2024$, every integer $k$ has rank $k + 44$. Solving $k + 44 = 2020$ gives $k = 1976$, the $2020$-th entry. The $2021$-st is $1977$. Median $= (1976 + 1977)/2 = \mathbf{1976.5}$, answer (C).
More like this
Same archetype — closest grade level first.