AMC 8 · 2025 · #6

Grade 4 number-theory

Archetype Small-Domain Constrained Search

multiplesmodular-arithmeticdivisibility-rules systematic-enumerationmodular-arithmetic ↑ Prerequisites: multi-digit-arithmeticmultiples

📏 Short solution 💡 2 insights

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Problem

Sekou writes the numbers $15, 16, 17, 18, 19.$ After he erases one of his numbers, the sum of the remaining four numbers is a multiple of $4.$ Which number did he erase?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Sekou wrote down the five consecutive whole numbers $15, 16, 17, 18, 19$. He erased exactly one of them, and the four numbers that are left add up to a multiple of $4$. Which of the five numbers did he erase?

Givens: The five starting numbers are $15, 16, 17, 18, 19$; Exactly one number is erased; The sum of the four remaining numbers is a multiple of $4$; Answer choices: (A) $15$, (B) $16$, (C) $17$, (D) $18$, (E) $19$

Unknowns: Which of the five numbers was erased

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List

There are only five candidates — the five answer choices themselves. That is a textbook setup for Tool #3 (Eliminate Possibilities): test each choice against the rule "sum of the remaining four is a multiple of $4$" and keep the one that survives. Tool #2 (Systematic List) keeps the bookkeeping tidy: first compute the total once, then list "$85$ minus each candidate" in order so no case is missed or double-counted. We deliberately avoid Tool #13 (Algebra) or modular-arithmetic shortcuts — they work, but a 4th-grader can solve this with addition and a divisibility check, which is the whole point.

Execute — Answer: C

#2 Make a Systematic List 4.NBT.B.4 Step 1

Add up all five numbers once.
Pairing the ends keeps the arithmetic easy: $(15+19) + (16+18) + 17 = 34 + 34 + 17 = 85$.

$$15 + 16 + 17 + 18 + 19 = 85$$

💡 Adding five two-digit numbers is exactly the Grade 4 "fluently add multi-digit whole numbers" skill — no shortcut needed.

#2 Make a Systematic List 4.NBT.B.4 Step 2

If the erased number is $x$, the sum of what is left is $85 - x$.
Make a systematic list of $85 - x$ for each of the five answer choices, in order, so every case is checked.

$$\begin{aligned} \text{(A) } 85-15 &= 70 \\ \text{(B) } 85-16 &= 69 \\ \text{(C) } 85-17 &= 68 \\ \text{(D) } 85-18 &= 67 \\ \text{(E) } 85-19 &= 66 \end{aligned}$$

💡 A clean ordered list of five subtractions makes sure no candidate is skipped — same Grade 4 add/subtract fluency.

#3 Eliminate Possibilities 4.OA.B.4 Step 3

Now eliminate each candidate whose remaining sum is NOT a multiple of $4$.
Test divisibility by $4$ for each value: $70 = 4 \times 17 + 2$ (no), $69$ is odd (no), $68 = 4 \times 17$ (yes), $67$ is odd (no), $66 = 4 \times 16 + 2$ (no).
Only choice (C) survives.

$$68 = 4 \times 17 \;\checkmark \qquad 70, 69, 67, 66 \text{ are not multiples of } 4$$

💡 Checking whether a number is a multiple of $4$ is exactly the Grade 4 "factors and multiples" skill — count by $4$s or divide and look for remainder $0$.

#3 Eliminate Possibilities 4.OA.B.4 Step 4

The only candidate that leaves a remaining sum that is a multiple of $4$ is $17$, so Sekou erased $17$ and the answer is (C).

$$\text{Erased number} = 17 \;\Rightarrow\; \textbf{(C)}$$

💡 Tool #3 says: when exactly one choice survives every test, that is the answer.

[1] #2 4.NBT.B.4 Add up all five numbers once. Pairing the ends keeps the arithmetic easy: $(15+1

[2] #2 4.NBT.B.4 If the erased number is $x$, the sum of what is left is $85 - x$. Make a systema

[3] #3 4.OA.B.4 Now eliminate each candidate whose remaining sum is NOT a multiple of $4$. Test

[4] #3 4.OA.B.4 The only candidate that leaves a remaining sum that is a multiple of $4$ is $17$

Review

Reasonableness: Quick sanity check: $68 \div 4 = 17$ with no remainder, so $68$ really is a multiple of $4$ and erasing $17$ works. Notice also that $85$ leaves a remainder of $1$ when divided by $4$, and among $15, 16, 17, 18, 19$ exactly one number ($17$) also leaves remainder $1$. That uniqueness explains why the problem has a single answer.

Alternative: Tool #5 (Look for a Pattern) gives a faster mental shortcut. Look at each starting number's remainder when divided by $4$: $15\to3,\; 16\to0,\; 17\to1,\; 18\to2,\; 19\to3$. The remainders sum to $3+0+1+2+3 = 9$, and $9$ leaves remainder $1$ when divided by $4$. So the number we erase must itself have remainder $1$ — and the only such number is $17$.

CCSS standards used (min grade 4)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding $15+16+17+18+19 = 85$ and then subtracting each candidate from $85$ to get the five possible remaining sums.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Checking which of $70, 69, 68, 67, 66$ is a multiple of $4$, which is the rule that decides the answer.)

⭐ This AMC 8 problem only needs Grade 4 addition and multiples-of-4 checking you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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