AMC 8 · 2016 · #5

Grade 4 number-theory

Archetype Small-Domain Constrained Search

modular-arithmeticdivisibility-rulesplace-value systematic-enumerationdigit-constraints ↑ Prerequisites: place-valuedivisibility-rules

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Problem

The number $N$ is a two-digit number.

• When $N$ is divided by $9$ , the remainder is $1$ .

• When $N$ is divided by $10$ , the remainder is $3$ .

What is the remainder when $N$ is divided by $11$ ?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: $N$ is a two-digit number. Dividing $N$ by $9$ leaves remainder $1$, and dividing $N$ by $10$ leaves remainder $3$. What is the remainder when $N$ is divided by $11$?

Givens: $N$ is a two-digit whole number ($10 \le N \le 99$); $N \div 9$ leaves remainder $1$; $N \div 10$ leaves remainder $3$; Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $5$, (E) $7$

Unknowns: The remainder when $N$ is divided by $11$

Understand

Restated: $N$ is a two-digit number. Dividing $N$ by $9$ leaves remainder $1$, and dividing $N$ by $10$ leaves remainder $3$. What is the remainder when $N$ is divided by $11$?

Givens: $N$ is a two-digit whole number ($10 \le N \le 99$); $N \div 9$ leaves remainder $1$; $N \div 10$ leaves remainder $3$; Answer choices: (A) $0$, (B) $2$, (C) $4$, (D) $5$, (E) $7$

Plan

Primary tool: #2 Make a Systematic List

Secondary: #3 Eliminate Possibilities, #5 Look for a Pattern

The "divide by $10$, remainder $3$" clue is very tight: it forces the units digit to be $3$, so only nine candidates exist ($13, 23, 33, \ldots, 93$). Tool #2 (Systematic List) writes them out in order. Tool #3 (Eliminate) then applies the second condition — divide by $9$, remainder $1$ — to throw out the candidates that fail. To check divisibility-by-$9$ quickly, we use Tool #5 (Look for a Pattern): a number's remainder mod $9$ equals the digit-sum's remainder mod $9$, so we just need the digit sum to leave remainder $1$. One candidate survives. Then a single division by $11$ finishes the problem — no algebra needed.

Execute — Answer: E

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

Use the mod-$10$ clue to list all candidates.
A number leaves remainder $3$ when divided by $10$ exactly when its units digit is $3$.
The two-digit numbers with units digit $3$ are exactly nine: $13, 23, 33, 43, 53, 63, 73, 83, 93$.

$$N \in \{13,\, 23,\, 33,\, 43,\, 53,\, 63,\, 73,\, 83,\, 93\}$$

💡 Reading the units digit straight off a number is a Grade 4 place-value move — the units digit IS the remainder when you divide by $10$.

#5 Look for a Pattern 4.OA.B.4 Step 2

Recall the divisibility-by-$9$ pattern.
For any whole number, the remainder when dividing by $9$ equals the remainder when dividing its digit sum by $9$.
So instead of dividing each candidate by $9$, we just add its digits and check whether that sum leaves remainder $1$.

$$N \bmod 9 = (\text{sum of digits of } N) \bmod 9$$

💡 Spotting the digit-sum shortcut is a Grade 4 multiples-and-divisibility pattern that saves nine long divisions.

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

Apply the digit-sum check to each candidate and eliminate the ones whose digit sum does not leave remainder $1$ when divided by $9$.
The digit sums are $4, 5, 6, 7, 8, 9, 10, 11, 12$, with remainders mod $9$ of $4, 5, 6, 7, 8, 0, 1, 2, 3$.
Only one candidate gives remainder $1$: $N = 73$.

$$\begin{array}{c|c|c} N & \text{digit sum} & \bmod 9 \\ \hline 13 & 4 & 4 \\ 23 & 5 & 5 \\ 33 & 6 & 6 \\ 43 & 7 & 7 \\ 53 & 8 & 8 \\ 63 & 9 & 0 \\ \mathbf{73} & \mathbf{10} & \mathbf{1} \\ 83 & 11 & 2 \\ 93 & 12 & 3 \end{array}$$

💡 Sweeping through a short candidate list and crossing off the failures is the textbook Grade 4 "reason about results" elimination move.

#2 Make a Systematic List 4.NBT.B.6 Step 4

Now divide $N = 73$ by $11$ to find the requested remainder.
Since $11 \times 6 = 66$ and $11 \times 7 = 77 > 73$, the quotient is $6$ and the remainder is $73 - 66 = 7$.

$$73 = 11 \times 6 + 7 \;\Rightarrow\; 73 \bmod 11 = 7 \;\Rightarrow\; \textbf{(E)}$$

💡 Dividing a two-digit number by a one- or two-digit number with a remainder is exactly the Grade 4 long-division standard.

[1] #2 4.NBT.A.2 Use the mod-$10$ clue to list all candidates. A number leaves remainder $3$ when

[2] #5 4.OA.B.4 Recall the divisibility-by-$9$ pattern. For any whole number, the remainder when

[3] #3 4.OA.A.3 Apply the digit-sum check to each candidate and eliminate the ones whose digit s

[4] #2 4.NBT.B.6 Now divide $N = 73$ by $11$ to find the requested remainder. Since $11 \times 6

Review

Reasonableness: Verify $N = 73$ against both original conditions: $73 = 9 \times 8 + 1$ ✓ (remainder $1$ when divided by $9$), and $73 = 10 \times 7 + 3$ ✓ (remainder $3$ when divided by $10$). Both check, so $N = 73$ is correct, and $73 \div 11 = 6$ remainder $7$ matches answer choice (E). The remainder $7$ is in $\{0, 1, \ldots, 10\}$ as a mod-$11$ remainder must be — magnitude is sensible.

Alternative: Tool #6 (Guess and Check) on the mod-$9$ side instead. Numbers leaving remainder $1$ when divided by $9$ are $10, 19, 28, 37, 46, 55, 64, 73, 82, 91$. From that list, pick the ones ending in digit $3$ — only $73$ qualifies. Same answer $N = 73$, then $73 \bmod 11 = 7 \Rightarrow$ (E). Either list-and-filter direction works; both are Tool #2 + Tool #3 combinations.

CCSS standards used (min grade 4)

4.NBT.A.2 Read and write multi-digit whole numbers using base-ten place value (Recognizing that "remainder $3$ when divided by $10$" means the units digit is $3$, which collapses $90$ two-digit numbers to just $9$ candidates.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Using the digit-sum divisibility pattern for $9$ ($N \bmod 9$ equals the digit sum $\bmod 9$) to test the second condition without doing nine long divisions.)
4.OA.A.3 Solve multi-step word problems using the four operations with whole numbers (Sweeping through the nine candidates, applying the mod-$9$ test to each, and eliminating until exactly one ($N = 73$) remains.)
4.NBT.B.6 Find whole-number quotients and remainders with multi-digit dividends (Computing $73 \div 11 = 6$ remainder $7$ to obtain the final answer.)

⭐ This AMC 8 problem only needs Grade 4 ideas — units-digit place value, the digit-sum rule for $9$, and basic division with remainders — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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