AMC 10 · 2020 · #11

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangeperfect-squaressystematic-enumerationfloor-function easier-related-problemidentify-subproblems ↑ Prerequisites: mean-median-mode-rangeperfect-squares

📏 Long solution 💡 3 insights

Problem

What is the median of the following list of $4040$ numbers $?$
$1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2$

Pick an answer.

(A)

1974.5

(B)

1975.5

(C)

1976.5

(D)

1977.5

(E)

1978.5

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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

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

#9 Solve an Easier Related Problem 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.

$$\text{sorted: } 1, 1, 2, 3, \mathbf{3}, \mathbf{4}, 4, 9, 16 \;\;\Rightarrow\;\; \text{median} = 3.5$$

💡 Grade 6 median: shrink the problem to a list you can sort by hand.

#2 Make a Systematic List 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 $44^2 = 1936 \le 2020 < 2025 = 45^2$).

$$44 \times 44 = 1936 \le 2020 < 2025 = 45 \times 45$$

💡 Grade 5 multi-digit multiplication: $44 \times 44$ and $45 \times 45$ pin down the squares that fit.

#5 Look for a Pattern 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 $k + 44$.

$$\text{rank}(k) = k + 44 \quad \text{for } 1936 \le k \le 2024$$

💡 Grade 4 pattern rule: each integer in that band picks up the same $44$ extra entries from squares.

#2 Make a Systematic List 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 $1976$.

$$\text{rank}(1976) = 1976 + 44 = 2020 \;\Rightarrow\; S_{2020} = 1976$$

💡 Grade 4 word-problem subtraction: $2020 - 44 = 1976$ undoes the rank shift.

#2 Make a Systematic List 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 $S_{2021} = 1977$.

$$S_{2021} = 1977 \;\; (\text{since } 1977 < 2025 = 45^2)$$

💡 Grade 4 comparing whole numbers: $1977 < 2025$, so $1977$ comes first.

#5 Look for a Pattern 6.SP.A.3 Step 6

Average the two middle terms to get the median.
$(1976 + 1977) / 2 = 3953/2 = 1976.5$.

$$\text{median} = \frac{1976 + 1977}{2} = 1976.5$$

💡 Grade 6 median: average the two middle entries of an even-length sorted list.

#3 Eliminate Possibilities 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.

$$1976.5 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 4 comparison: only one choice equals $1976.5$.

[1] #9 6.SP.A.3 Try a tiny version first: $N=4$. The list is $\{1, 2, 3, 4, 1, 4, 9, 16\}$, whic

[2] #2 5.NBT.B.5 Generalize the lesson from the small case. To find the rank of a value $k$ in th

[3] #5 4.OA.C.5 Build the counting pattern. For any integer $k$ between $44^2 = 1936$ and $45^2

[4] #2 4.OA.A.3 Find the $2020$-th term. Solve $k + 44 = 2020$, so $k = 1976$. Check: $1976$ is

[5] #2 4.NBT.A.2 Find the $2021$-st term. The next candidate is the next integer $1977$ (still le

[6] #5 6.SP.A.3 Average the two middle terms to get the median. $(1976 + 1977) / 2 = 3953/2 = 19

[7] #3 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.2 Read 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.3 Solve 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.5 Generate 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.5 Fluently multiply multi-digit whole numbers (Computing $44 \times 44 = 1936$ and $45 \times 45 = 2025$ to bracket the squares $\le 2020$.)
6.SP.A.3 Recognize 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).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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