AMC 8 · 2018 · #7
Grade 4 number-theoryProblem
The 5-digit number 2 0 1 8 U is divisible by 9. What is the remainder when this number is divided by 8?
Pick an answer.
AMC 8 2018 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: We are given a $5$-digit number whose first four digits are $2, 0, 1, 8$ and whose last digit is an unknown digit $U$. We are told the whole number $2018U$ is divisible by $9$. We need to find the remainder when this $5$-digit number is divided by $8$.
Givens: The number has the form $\overline{2018U}$ where $U$ is a single digit from $0$ to $9$; $\overline{2018U}$ is divisible by $9$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The digit $U$; The remainder when $\overline{2018U}$ is divided by $8$
Understand
Restated: We are given a $5$-digit number whose first four digits are $2, 0, 1, 8$ and whose last digit is an unknown digit $U$. We are told the whole number $2018U$ is divisible by $9$. We need to find the remainder when this $5$-digit number is divided by $8$.
Givens: The number has the form $\overline{2018U}$ where $U$ is a single digit from $0$ to $9$; $\overline{2018U}$ is divisible by $9$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #6 Guess and Check
The question hides two clean sub-questions (Tool #7): (a) use the divisibility-by-$9$ clue to figure out the digit $U$, and (b) once the number is known, compute its remainder when divided by $8$. Sub-question (a) is exactly the digit-sum pattern for $9$ (Tool #5: a number is divisible by $9$ when its digits add to a multiple of $9$); since $U$ ranges over only $10$ values, Tool #6 (Guess and Check) also works in seconds. Sub-question (b) is a single division with remainder, no algebra needed.
Execute — Answer: B
4.NBT.B.4 Step 1 - Sub-problem A: find $U$ using the divisibility rule for $9$.
- A whole number is a multiple of $9$ exactly when the sum of its digits is a multiple of $9$.
- Add the known digits of $\overline{2018U}$ and leave $U$ as an unknown addend.
💡 Adding the four known digits is a Grade 4 multi-digit addition fact ($2+0+1+8 = 11$).
4.OA.B.4 Step 2 - Find which single-digit $U$ makes $11 + U$ a multiple of $9$.
- Since $U$ is one digit, $11 + U$ lies between $11$ and $20$, and the only multiple of $9$ in that range is $18$.
- So $11 + U = 18$, giving $U = 7$.
- (A quick Tool #6 Guess-and-Check across $U = 0, 1, 2, \ldots, 9$ confirms the same: only $U = 7$ produces a digit sum divisible by $9$.)
💡 Checking which sum is a multiple of $9$ is a Grade 4 multiples-of-a-number task.
4.OA.B.4 Step 3 - Sub-problem B: compute the remainder when $20187$ is divided by $8$.
- We can take a shortcut from the pattern that $1000 = 8 \times 125$ is a multiple of $8$, so $20000 = 20 \times 1000$ is also a multiple of $8$.
- That means the remainder of $20187$ when divided by $8$ equals the remainder of just its last three digits, $187$, when divided by $8$.
💡 Spotting that multiples of $1000$ are also multiples of $8$ is a Grade 4 multiples observation.
4.NBT.B.6 Step 4 - Divide $187$ by $8$ with remainder.
- The largest multiple of $8$ at most $187$ is $8 \times 23 = 184$, leaving $187 - 184 = 3$.
- So $187 = 8 \times 23 + 3$, and the remainder is $3$.
💡 Finding the quotient $23$ and remainder $3$ from a three-digit dividend is exactly the Grade 4 division-with-remainder standard.
4.NBT.B.6 Step 5 - Combine: $20187 \bmod 8 = 187 \bmod 8 = 3$.
- The remainder is $3$, which matches answer choice (B).
💡 Stitching the two sub-answers together is the Tool #7 "combine subproblems" move.
4.NBT.B.4 Sub-problem A: find $U$ using the divisibility rule for $9$. A whole number is a 4.OA.B.4 Find which single-digit $U$ makes $11 + U$ a multiple of $9$. Since $U$ is one d 4.OA.B.4 Sub-problem B: compute the remainder when $20187$ is divided by $8$. We can take 4.NBT.B.6 Divide $187$ by $8$ with remainder. The largest multiple of $8$ at most $187$ is 4.NBT.B.6 Combine: $20187 \bmod 8 = 187 \bmod 8 = 3$. The remainder is $3$, which matches Review
Reasonableness: A remainder when dividing by $8$ must be a whole number from $0$ to $7$, and $3$ fits. Double-check by direct division: $20187 \div 8 = 2523$ with remainder $3$, since $8 \times 2523 = 20184$ and $20187 - 20184 = 3$. We can also verify $U = 7$ is correct: $20187 \div 9 = 2243$ exactly ($9 \times 2243 = 20187$), so the original divisibility condition holds. Both halves check out, so the answer (B) $3$ is solid.
Alternative: Tool #3 (Eliminate Possibilities) plus direct division. Once $U = 7$ is found, just long-divide $20187$ by $8$: the remainder must be one of (A) $1$, (B) $3$, (C) $5$, (D) $6$, (E) $7$, and the long division gives $3$ directly, eliminating all other choices in one shot.
CCSS standards used (min grade 4)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the four known digits $2 + 0 + 1 + 8 = 11$ to set up the digit-sum equation $11 + U$ for the divisibility-by-$9$ test.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing which value of $11 + U$ is a multiple of $9$ (giving $U = 7$), and observing that $1000$ is a multiple of $8$ so any multiple of $1000$ is too.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $187$ by $8$ to get the quotient $23$ and remainder $3$, which is the final answer.)
⭐ This AMC 8 problem only needs Grade 4 divisibility rules and division-with-remainder you already know!
⭐ This AMC 8 problem only needs Grade 4 divisibility rules and division-with-remainder you already know!
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