AMC 10 · 2024 · #1
Easy mode Grade 4Problem
Imagine a long line of people standing side by side, from left to right.
Pick out one person in this line. Count from the left, and this person is the 1013th. Count from the right, and the same person is the 1010th.
How many people are in the whole line?
Pick an answer.
AMC 10 2024 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: 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?
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) $2025$
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
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.
💡 Kindergarten position words — left of, beside, right of — are exactly what the problem describes, and a quick sketch turns the words into something countable.
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.
💡 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.
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.
💡 Same idea on the other side: the $1010$th-from-right has $1009$ people to their right.
4.NBT.B.4 Step 4 Add the three pieces — left, the person, right — to get the total length of the line.
💡 Combining three disjoint groups is the addition principle of counting; the four-digit sum is exactly the Grade 4 multi-digit addition fluency.
K.G.A.1 Draw the line as three labeled pieces: the people to the left, the one specific 2.OA.A.1 Count the people to the left. Being the $1013$th from the left means $1012$ peop 2.OA.A.1 Count the people to the right. Being the $1010$th from the right means $1009$ pe 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.1Describe 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.1Solve 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.4Fluently 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!
⭐ This AMC 10 problem only needs Grade 4 multi-digit addition (and one careful subtract-$1$ from position counting) that you already know!
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