AMC 10 · 2024 · #4

Grade 6 arithmeticnumber-theory

Archetype Small-Domain Constrained Search

multi-digit-arithmeticbound-inequality-then-enumerateinterval-arithmetic bound-inequality-then-enumerateoptimization-counting ↑ Prerequisites: multi-digit-arithmeticmultiples

📏 Short solution 💡 2 insights

Problem

The number $2024$ is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?

$\textbf{(A) }20\qquad\textbf{(B) }21\qquad\textbf{(C) }22\qquad\textbf{(D) }23\qquad\textbf{(E) }24$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Write $2024$ as a sum of two-digit numbers (repeats allowed). What is the smallest possible count of terms?

Givens: The target sum is $2024$; Every term must be a two-digit positive integer, i.e. between $10$ and $99$; Terms may repeat; Answer choices: (A) $20$, (B) $21$, (C) $22$, (D) $23$, (E) $24$

Unknowns: The minimum number of two-digit terms needed

Understand

Restated: Write $2024$ as a sum of two-digit numbers (repeats allowed). What is the smallest possible count of terms?

Givens: The target sum is $2024$; Every term must be a two-digit positive integer, i.e. between $10$ and $99$; Terms may repeat; Answer choices: (A) $20$, (B) $21$, (C) $22$, (D) $23$, (E) $24$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #6 Guess and Check

An "at least" minimum problem really hides two questions, so Tool #7 (Identify Subproblems) splits the work cleanly. Subproblem A: how few terms could possibly work? Each term is at most $99$, so $n$ terms can reach at most $99n$ — that gives a lower bound on $n$. Subproblem B: can that lower bound actually be achieved with valid two-digit numbers? We pair this with Tool #6 (Guess and Check) to test the boundary cases $n = 20$ and $n = 21$ — one fails, one works, and that pins down the answer.

Execute — Answer: B

#7 Identify Subproblems 6.EE.B.8 Step 1

Subproblem A — find a lower bound.
If we use $n$ two-digit numbers, each is at most $99$, so the total cannot exceed $99n$.
To reach $2024$ we need $99n \ge 2024$.

$$99n \ge 2024 \;\Rightarrow\; n \ge \dfrac{2024}{99}$$

💡 Writing the constraint as an inequality is exactly the Grade 6 move "$n \ge c$": $n$ has to be at least as big as $\tfrac{2024}{99}$.

#6 Guess and Check 6.NS.B.2 Step 2

Divide to turn the inequality into a number.
$99 \times 20 = 1980 < 2024$, so $n = 20$ is too small.
Try $n = 21$: $99 \times 21 = 2079 \ge 2024$, so $n = 21$ clears the bound.

$$99 \times 20 = 1980 \;\;\text{(short)}, \quad 99 \times 21 = 2079 \;\;\text{(enough)}$$

💡 Grade 6 fluent multi-digit arithmetic: just multiply $99$ by $20$ and $21$ to see which one first crosses $2024$.

#7 Identify Subproblems 6.NS.B.2 Step 3

Subproblem B — show $n = 21$ is actually reachable.
We need an honest sum of $21$ two-digit numbers equal to $2024$.
Write $2024 = 99 \times 20 + r$ and solve for $r$.

$$2024 - 20 \times 99 = 2024 - 1980 = 44$$

💡 After using twenty $99$s, we still need $44$ more — and $44$ is itself a two-digit number, which is the perfect $21$st term.

#6 Guess and Check 6.EE.B.5 Step 4

Assemble the example and confirm.
Twenty $99$s plus one $44$ uses exactly $21$ two-digit numbers and sums to $2024$.
Combined with the lower bound from Step 2, the minimum is $21$.

$$\underbrace{99 + 99 + \cdots + 99}_{20 \text{ terms}} + 44 = 1980 + 44 = 2024 \;\Rightarrow\; \textbf{(B)}$$

💡 The candidate value $n = 21$ passes the check (a valid sum exists), so the inequality bound is tight — that's how Grade 6 "is this $n$ a solution?" reasoning closes the problem.

[1] #7 6.EE.B.8 Subproblem A — find a lower bound. If we use $n$ two-digit numbers, each is at m

[2] #6 6.NS.B.2 Divide to turn the inequality into a number. $99 \times 20 = 1980 < 2024$, so $n

[3] #7 6.NS.B.2 Subproblem B — show $n = 21$ is actually reachable. We need an honest sum of $21

[4] #6 6.EE.B.5 Assemble the example and confirm. Twenty $99$s plus one $44$ uses exactly $21$ t

Review

Reasonableness: Sanity-check the boundary: with $20$ terms the largest possible total is $20 \times 99 = 1980$, which is $44$ short of $2024$. There is no way to patch this with two-digit numbers because every term is already at its ceiling, so $20$ is genuinely impossible. With $21$ terms we exhibited a concrete sum equal to $2024$, so $21$ is genuinely possible. The answer must therefore be $21$, matching (B). The construction also uses only two distinct values ($99$ and $44$), confirming the recipe is straightforward.

Alternative: Tool #9 (Solve an Easier Related Problem): scale the question down. "Write $20$ as a sum of one-digit numbers ($1$ through $9$) using the fewest terms." The biggest one-digit number is $9$, and $9 \times 2 = 18 < 20$, while $9 \times 3 = 27 \ge 20$, so the minimum is $3$ (e.g., $9 + 9 + 2$). Same recipe: ceiling-divide the target by the maximum allowed term. Applying it to the original gives $\lceil 2024 / 99 \rceil = 21$.

CCSS standards used (min grade 6)

6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Translating "each term is at most $99$" into the inequality $99n \ge 2024$, which encodes the lower bound on the number of terms.)
6.NS.B.2 Fluently divide multi-digit numbers using the standard algorithm (Computing $99 \times 20 = 1980$ and $99 \times 21 = 2079$ to bracket $2024$, and computing $2024 - 1980 = 44$ to find the leftover term.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Checking that $n = 21$ is an actual solution (the inequality bound is achievable), not just an arithmetic lower bound.)

⭐ An "as few as possible" problem becomes easy when you ask "as big as possible per piece?" first — the AMC 10's opening problems often reward this Grade 6 inequality + check move.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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