AMC 8 · 2018 · #7
Easy mode Grade 4Problem
Picture a 5-digit number that starts with the digits 2, 0, 1, 8, and then has one more digit at the end. We will call that last digit U.
So the number looks like 2018U.
Here is what we know: this number is divisible by 9. That fact tells you what U has to be.
Once you figure out U, you have the full 5-digit number. The question asks: what is the remainder when this 5-digit 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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