AMC 10 · 2024 · #1

Grade 4 arithmetic

Archetype Arithmetic Expression Decomposition

multi-digit-arithmeticpattern-recognitioncomplementary-counting identify-subproblemseasier-related-problem ↑ Prerequisites: multi-digit-arithmeticorder-of-operations

📏 Short solution 💡 2 insights

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Problem

In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?

$\textbf{(A) } 2021 \qquad\textbf{(B) } 2022 \qquad\textbf{(C) } 2023 \qquad\textbf{(D) } 2024 \qquad\textbf{(E) } 2025$

Modified Problem in Certain China Testpapers

In a long line of people arranged left to right, the 1015th person from the left is also the 1010th person from the right. How many people are in line?

$\textbf{(A) } 2021 \qquad\textbf{(B) } 2022 \qquad\textbf{(C) } 2023 \qquad\textbf{(D) } 2024 \qquad\textbf{(E) } 2025$

Solution for Certain China Testpapers

If the person is the 1015th from the left, that means there is 1014 people to their left.
If the person is the 1010th from the right, that means there is 1009 people to their right.
Therefore, there are $1014 + 1 + 1009 = \boxed{\textbf{(D) } 2024}$ people in line.

~Aray10 (Main Solution) and RULE101 (Modifications for certain China test papers)

Pick an answer.

(A)

2021

(B)

2022

(C)

2023

(D)

2024

(E)

2025

View mode:

Toolkit + CCSS Solution

Understand

Restated: One person stands in a single line. Counting from the left, they are the $1013$th person; counting from the right, they are the $1010$th person. How many people are in the whole line?

Givens: Position from the left: $1013$th; Position from the right: $1010$th; Everyone, including this person, stands in one continuous line; Answer choices: (A) $2021$, (B) $2022$, (C) $2023$, (D) $2024$, (E) $2025$

Unknowns: The total number of people in the line

Understand

Restated: One person stands in a single line. Counting from the left, they are the $1013$th person; counting from the right, they are the $1010$th person. How many people are in the whole line?

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The problem is a position-on-a-line picture: one fixed person, some people to the left, some people to the right. Tool #1 (Draw a Diagram) makes that picture explicit — sketch L L L ... L | P | R R R ... R and the count becomes visible. Tool #7 (Identify Subproblems) cleanly splits the line into three independent pieces — left of the person, the person themselves, right of the person — so we can count each piece and add.

Execute — Answer: B

#1 Draw a Diagram K.G.A.1 Step 1

Draw the line as three labeled pieces: the people to the left, the one specific person in the middle, and the people to the right.
The sketch makes it obvious that adding the three counts gives the total.

$$\underbrace{L \; L \; \cdots \; L}_{\text{left of P}} \;\big|\; P \;\big|\; \underbrace{R \; R \; \cdots \; R}_{\text{right of P}}$$

💡 Kindergarten position words — left of, beside, right of — are exactly what the problem describes, and a quick sketch turns the words into something countable.

#7 Identify Subproblems 2.OA.A.1 Step 2

Count the people to the left.
Being the $1013$th from the left means $1012$ people stand before this person on the left side.

$$\text{left of P} = 1013 - 1 = 1012 \text{ people}$$

💡 If you are number $1013$ in a counted-from-$1$ line, $1012$ people came before you — a one-step word-problem subtraction within Grade 2 reach.

#7 Identify Subproblems 2.OA.A.1 Step 3

Count the people to the right.
Being the $1010$th from the right means $1009$ people stand after this person on the right side.

$$\text{right of P} = 1010 - 1 = 1009 \text{ people}$$

💡 Same idea on the other side: the $1010$th-from-right has $1009$ people to their right.

#7 Identify Subproblems 4.NBT.B.4 Step 4

Add the three pieces — left, the person, right — to get the total length of the line.

$$1012 + 1 + 1009 = 2022 \;\Rightarrow\; \textbf{(B)}$$

💡 Combining three disjoint groups is the addition principle of counting; the four-digit sum is exactly the Grade 4 multi-digit addition fluency.

[1] #1 K.G.A.1 Draw the line as three labeled pieces: the people to the left, the one specific

[2] #7 2.OA.A.1 Count the people to the left. Being the $1013$th from the left means $1012$ peop

[3] #7 2.OA.A.1 Count the people to the right. Being the $1010$th from the right means $1009$ pe

[4] #7 4.NBT.B.4 Add the three pieces — left, the person, right — to get the total length of the

Review

Reasonableness: Sanity-check with the shortcut formula: $\text{total} = (\text{from left}) + (\text{from right}) - 1 = 1013 + 1010 - 1 = 2022$. The $-1$ removes the double-count of the one person who is named from both sides. Same answer (B). Also, $1013 + 1010 = 2023$ is just barely above $2022$, so an answer near $2022$ is exactly what we expect.

Alternative: Tool #9 (Solve an Easier Related Problem): test on a tiny line. If someone is $3$rd from the left and $4$th from the right, draw L L P R R R — that's $6$ people total, and $3 + 4 - 1 = 6$. The pattern "add the two positions and subtract $1$" carries straight over to $1013 + 1010 - 1 = 2022$.

CCSS standards used (min grade 4)

K.G.A.1 Describe positions of objects using above, below, beside, in front of (Using position language (left of, right of, beside) to organize the line into three labeled regions before counting.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Translating "the $n$th from one end" into "$n - 1$ people stand on that side" — a one-step subtraction word problem.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding the three four-digit-range counts $1012 + 1 + 1009 = 2022$ to get the total number of people.)

⭐ This AMC 10 problem only needs Grade 4 multi-digit addition (and one careful subtract-$1$ from position counting) that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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