AMC 8 · 2024 · #4

Easy mode Grade 3

Problem

Imagine Yunji is adding up the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ .

By mistake, she skipped one of those nine numbers and added the rest.

The wrong total she got happens to be a perfect square (a number like $1, 4, 9, 16, 25, 36, \dots$ that is some whole number times itself).

Which number from $1$ through $9$ did Yunji skip?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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.

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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

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 Make a Systematic List 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 $40 + 5 = 45$.

$$1+2+3+4+5+6+7+8+9 = (1+9)+(2+8)+(3+7)+(4+6)+5 = 10+10+10+10+5 = 45$$

💡 Adding nine single-digit numbers using a pairing trick is Grade 2 addition fluency within 100.

#6 Guess and Check 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).

$$x \in \{1, 2, \ldots, 9\} \;\Rightarrow\; 45 - 9 \le 45 - x \le 45 - 1 \;\Rightarrow\; 36 \le 45 - x \le 44$$

💡 Using a one-step subtraction within 100 to bound the possible values is a Grade 2 word-problem skill.

#2 Make a Systematic List 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 $36$.

$$1, 4, 9, 16, 25, \underbrace{36}_{\in [36,44]\ \checkmark}, 49, \ldots \;\Rightarrow\; \text{wrong sum} = 36 = 6^2$$

💡 Recognizing $6 \times 6 = 36$ as a perfect square uses Grade 3 single-digit multiplication fluency within 100.

#11 Work Backwards 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)**.

$$45 - x = 36 \;\Rightarrow\; x = 45 - 36 = 9 \;\Rightarrow\; \textbf{(E)}$$

💡 Solving "what do I subtract from 45 to get 36?" is a Grade 2 unknown-addend / subtraction-within-100 task.

[1] #2 2.NBT.B.5 First, find the correct sum of $1$ through $9$. Pair the numbers from the outsid

[2] #6 2.OA.A.1 Let $x$ be the number Yunji left out. Her wrong sum is $45 - x$. Since $x$ is on

[3] #2 3.OA.C.7 Now list perfect squares in order and check which (if any) land in $[36, 44]$: $

[4] #11 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.5 Fluently add and subtract within 100 (Adding $1+2+\cdots+9 = 45$ using the pairing trick.)
2.OA.A.1 Solve 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.7 Fluently 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 3)

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