AMC 8 · 2024 · #4
Grade 3 geometry-2dProblem
When Yunji added all the integers from 1 to 9, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
Pick an answer.
AMC 8 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: Yunji meant to add every integer from $1$ to $9$, but she accidentally left one of them out. Her resulting (wrong) sum happens to be a perfect square. Among the answer choices (A)~(E), we must identify which integer she left out.
Givens: The intended addends are the nine integers $1, 2, 3, 4, 5, 6, 7, 8, 9$; Exactly one of these nine integers was omitted; The sum she actually computed is a perfect square (some integer squared); Five candidate missing numbers: (A) 5, (B) 6, (C) 7, (D) 8, (E) 9
Unknowns: The integer $x$ Yunji left out (a single value from $1$ through $9$)
Understand
Restated: Yunji meant to add every integer from $1$ to $9$, but she accidentally left one of them out. Her resulting (wrong) sum happens to be a perfect square. Among the answer choices (A)~(E), we must identify which integer she left out.
Givens: The intended addends are the nine integers $1, 2, 3, 4, 5, 6, 7, 8, 9$; Exactly one of these nine integers was omitted; The sum she actually computed is a perfect square (some integer squared); Five candidate missing numbers: (A) 5, (B) 6, (C) 7, (D) 8, (E) 9
Plan
Primary tool: #2 Make a Systematic List
Secondary: #6 Guess and Check, #11 Work Backwards
Everything in this problem lives in a tiny range — there are only 9 possible missing numbers and the wrong sum is squeezed into a short interval. So we just **list** the perfect squares in order ($1, 4, 9, 16, 25, 36, 49, \ldots$) until one falls in the allowed range (Tool #2). We can double-check by **guessing and checking** each answer choice (Tool #6). Once we know the wrong sum is 36, we **work backwards** from $45 - x = 36$ to recover $x$ (Tool #11). No algebraic machinery (Tool #13) is needed for an elementary student.
Execute — Answer: E
2.NBT.B.5 Step 1 - First, find the correct sum of $1$ through $9$.
- Pair the numbers from the outside in: $1+9=10$, $2+8=10$, $3+7=10$, $4+6=10$, with $5$ left over in the middle.
- That gives four 10s and a 5, so the total is $40 + 5 = 45$.
💡 Adding nine single-digit numbers using a pairing trick is Grade 2 addition fluency within 100.
2.OA.A.1 Step 2 - Let $x$ be the number Yunji left out.
- Her wrong sum is $45 - x$.
- Since $x$ is one of $1, 2, \ldots, 9$, the wrong sum is at most $45-1 = 44$ and at least $45-9 = 36$.
- So the wrong sum must lie in the range **36 to 44** (inclusive).
💡 Using a one-step subtraction within 100 to bound the possible values is a Grade 2 word-problem skill.
3.OA.C.7 Step 3 - Now list perfect squares in order and check which (if any) land in $[36, 44]$: $1^2=1$, $2^2=4$, $3^2=9$, $4^2=16$, $5^2=25$, $6^2=36$, $7^2=49$.
- Only $36$ fits — $25$ is too small and $49$ is too big.
- So the wrong sum must equal $36$.
💡 Recognizing $6 \times 6 = 36$ as a perfect square uses Grade 3 single-digit multiplication fluency within 100.
2.OA.A.1 Step 4 - Finally, work backwards from $45 - x = 36$ to find $x$.
- Subtracting in reverse, $x = 45 - 36 = 9$.
- So Yunji left out the number **9**, which is answer **(E)**.
💡 Solving "what do I subtract from 45 to get 36?" is a Grade 2 unknown-addend / subtraction-within-100 task.
2.NBT.B.5 First, find the correct sum of $1$ through $9$. Pair the numbers from the outsid 2.OA.A.1 Let $x$ be the number Yunji left out. Her wrong sum is $45 - x$. Since $x$ is on 3.OA.C.7 Now list perfect squares in order and check which (if any) land in $[36, 44]$: $ 2.OA.A.1 Finally, work backwards from $45 - x = 36$ to find $x$. Subtracting in reverse, Review
Reasonableness: Let's double-check directly. If Yunji left out 9, her sum is $1+2+3+4+5+6+7+8 = 36 = 6 \times 6 = 6^2$ — indeed a perfect square. The range $[36, 44]$ contains only one perfect square (36), so the answer is unique. We can also rule out the other choices by direct subtraction: $45-5=40$, $45-6=39$, $45-7=38$, $45-8=37$ — none of these is a perfect square. So answer (E) 9 is correct.
Alternative: Tool #3 (Eliminate Possibilities) gives an even shorter path on a multiple-choice problem: just subtract each candidate from 45 and check for a perfect square. $45-5=40$ (no), $45-6=39$ (no), $45-7=38$ (no), $45-8=37$ (no), $45-9=36 = 6^2$ ($\checkmark$). Answer (E) drops out immediately. Whenever the answer choices are concrete, plugging them in is the fastest AMC-8 strategy.
CCSS standards used (min grade 3)
2.NBT.B.5Fluently add and subtract within 100 (Adding $1+2+\cdots+9 = 45$ using the pairing trick.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Bounding the wrong sum to the range $[36, 44]$ and then solving $45 - x = 36$ for $x$ by working backwards.)3.OA.C.7Fluently multiply and divide within 100 (Listing the perfect squares $1, 4, 9, 16, 25, 36, 49$ and recognizing $36 = 6 \times 6$.)
⭐ This AMC 8 problem only needs the Grade 3 times-table fact $6 \times 6 = 36$ you already know!
⭐ This AMC 8 problem only needs the Grade 3 times-table fact $6 \times 6 = 36$ you already know!
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