AMC 8 · 2018 · #7

Easy mode Grade 4

Problem

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 $\underline{2}\,\underline{0}\,\underline{1}\,\underline{8}\,\underline{U}$ .

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 $8$ ?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#5 Look for a Pattern 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.

$$2 + 0 + 1 + 8 + U = 11 + U$$

💡 Adding the four known digits is a Grade 4 multi-digit addition fact ($2+0+1+8 = 11$).

#6 Guess and Check 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 $9$.)

$$11 + U = 18 \;\Rightarrow\; U = 7 \;\Rightarrow\; \overline{2018U} = 20187$$

💡 Checking which sum is a multiple of $9$ is a Grade 4 multiples-of-a-number task.

#5 Look for a Pattern 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 $8$.

$$20187 = 20000 + 187,\;\; 20000 = 8 \times 2500 \;\Rightarrow\; 20187 \bmod 8 = 187 \bmod 8$$

💡 Spotting that multiples of $1000$ are also multiples of $8$ is a Grade 4 multiples observation.

#7 Identify Subproblems 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 $3$.

$$187 = 8 \times 23 + 3 \;\Rightarrow\; 187 \bmod 8 = 3$$

💡 Finding the quotient $23$ and remainder $3$ from a three-digit dividend is exactly the Grade 4 division-with-remainder standard.

#7 Identify Subproblems 4.NBT.B.6 Step 5

Combine: $20187 \bmod 8 = 187 \bmod 8 = 3$.
The remainder is $3$, which matches answer choice (B).

$$20187 \bmod 8 = 3 \;\Rightarrow\; \textbf{(B)}$$

💡 Stitching the two sub-answers together is the Tool #7 "combine subproblems" move.

[1] #5 4.NBT.B.4 Sub-problem A: find $U$ using the divisibility rule for $9$. A whole number is a

[2] #6 4.OA.B.4 Find which single-digit $U$ makes $11 + U$ a multiple of $9$. Since $U$ is one d

[3] #5 4.OA.B.4 Sub-problem B: compute the remainder when $20187$ is divided by $8$. We can take

[4] #7 4.NBT.B.6 Divide $187$ by $8$ with remainder. The largest multiple of $8$ at most $187$ is

[5] #7 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.4 Fluently 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.4 Find 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.6 Find 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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