AMC 8 · 2025 · #6
Easy mode Grade 4Problem
Imagine Sekou writing five numbers in a row on a whiteboard: 15, 16, 17, 18, 19.
Then he picks one of the five and erases it. Four numbers are left on the board.
He adds those four leftover numbers together. The total turns out to be a multiple of 4 — in other words, it divides evenly by 4 with nothing left over.
Which one of the five numbers did Sekou erase?
Pick an answer.
AMC 8 2025 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: 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
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) $19$
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
4.NBT.B.4 Step 1 - Add up all five numbers once.
- Pairing the ends keeps the arithmetic easy: $(15+19) + (16+18) + 17 = 34 + 34 + 17 = 85$.
💡 Adding five two-digit numbers is exactly the Grade 4 "fluently add multi-digit whole numbers" skill — no shortcut needed.
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.
💡 A clean ordered list of five subtractions makes sure no candidate is skipped — same Grade 4 add/subtract fluency.
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.
💡 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 $0$.
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).
💡 Tool #3 says: when exactly one choice survives every test, that is the answer.
4.NBT.B.4 Add up all five numbers once. Pairing the ends keeps the arithmetic easy: $(15+1 4.NBT.B.4 If the erased number is $x$, the sum of what is left is $85 - x$. Make a systema 4.OA.B.4 Now eliminate each candidate whose remaining sum is NOT a multiple of $4$. Test 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.4Fluently 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.4Find 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!
⭐ This AMC 8 problem only needs Grade 4 addition and multiples-of-4 checking you already know!
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