AMC 10 · 2024 · #1
Grade 4 arithmeticPick an answer.
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.
Split the line into three
Left, the person, and right.
Kindergarten position words — left of, beside, right of — are exactly what the problem describes, and a quick sketch turns the words into something countable.
K.G.A.1Draw A DiagramCount the left side
There are 1012 on the left.
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.1Identify SubproblemsCount the right side
There are 1009 on the right.
Same idea on the other side: the 1010th-from-right has 1009 people to their right.
2.OA.A.1Identify SubproblemsAdd the three pieces
1012 plus 1 plus 1009 is 2022.
Combining three disjoint groups is the addition principle of counting; the four-digit sum is exactly the Grade 4 multi-digit addition fluency.
Combining three groups that share nobody is a plain addition.
▸ Why?
Each person falls into exactly one of the three groups, so nobody is counted twice.
▸ Why?
The line is exactly those three groups put together, so their counts fill it completely.
This AMC 12 problem only needs Grade 4 multi-digit addition (and one careful subtract-1 from position counting) that you already know!