AMC 8 · 2000 · #11

Grade 4 number-theory

Archetype Casework by Position / Vertex Type └ Sub-archetype Digit Place Position

divisibility-rulesdigit-constraintssystematic-enumeration caseworksystematic-enumerationdigit-constraints ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2000 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.

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Toolkit + CCSS Solution

Understand

Restated: The number $64$ is divisible by its units digit because $64 \div 4 = 16$. How many whole numbers strictly between $10$ and $50$ share this property — that is, the number is divisible by whichever digit happens to sit in its units place?

Givens: Range: whole numbers strictly between $10$ and $50$, i.e. $11, 12, 13, \dots, 49$; A number $N$ "has the property" when $N$ is divisible by its units digit $u$; $N$ is excluded automatically when $u = 0$, because division by $0$ is undefined; Answer choices: (A) $15$, (B) $16$, (C) $17$, (D) $18$, (E) $20$

Unknowns: The count of numbers in $\{11, 12, \dots, 49\}$ that are divisible by their units digit

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #3 Eliminate Possibilities

There are only $39$ numbers in the range, so a brute-force check is plausible — but smarter is Tool #2 (Make a Systematic List) grouped by the units digit $u$. Each $u$ from $1$ to $9$ gives at most $4$ candidates ($1u, 2u, 3u, 4u$), and the question "is $N$ divisible by $u$?" is identical for the whole group. Tool #3 (Eliminate Possibilities) then knocks out non-multiples within each group with a single divisibility rule. Splitting by units digit turns one $39$-item check into nine tiny checks of at most $4$ each — much easier to count without errors.

Execute — Answer: C

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

Set up the casework.
Fix the units digit $u$ from $1$ to $9$ (skip $u = 0$ since no number is divisible by $0$).
For each $u$, the candidates in the range are $1u, 2u, 3u, 4u$ — exactly four numbers — and the question is which of those four are multiples of $u$.

$$\text{Candidates for units digit } u: \;\{10+u,\; 20+u,\; 30+u,\; 40+u\}$$

💡 Grade 4 place value: a two-digit number is (tens digit) $\times 10 +$ (units digit). Fixing $u$ leaves only the tens digit to vary.

#3 Eliminate Possibilities 4.OA.B.4 Step 2

Handle the easy units digits — the ones where the divisibility rule guarantees all four candidates work.
For $u = 1$, every whole number is divisible by $1$.
For $u = 2$, every number ending in $2$ is even, so divisible by $2$.
For $u = 5$, every number ending in $5$ is divisible by $5$.

$$u=1: \{11,21,31,41\} \to 4. \quad u=2: \{12,22,32,42\} \to 4. \quad u=5: \{15,25,35,45\} \to 4.$$

💡 Three digits ($1, 2, 5$) come free because divisibility by them is built into how the number ends — Grade 4 divisibility rules.

#3 Eliminate Possibilities 3.OA.C.7 Step 3

Now the medium cases.
For $u = 4$, check each candidate: $14, 24, 34, 44$.
Dividing by $4$: $14 / 4$ is not whole, $24 / 4 = 6$, $34 / 4$ is not whole, $44 / 4 = 11$.
For $u = 3$: $13, 23, 33, 43$.
Only $33 = 3 \times 11$ works.
For $u = 6$: $16, 26, 36, 46$.
Only $36 = 6 \times 6$ works.
For $u = 8$: $18, 28, 38, 48$.
Only $48 = 8 \times 6$ works.

$$u=4: \{24,44\} \to 2. \quad u=3: \{33\} \to 1. \quad u=6: \{36\} \to 1. \quad u=8: \{48\} \to 1.$$

💡 Just multiplication-table recall: scan the multiples of $u$ up to about $50$ and keep the ones whose units digit really is $u$.

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

The hard-but-empty cases.
For $u = 7$, the multiples of $7$ are $7, 14, 21, 28, 35, 42, 49$ — none ends in $7$ inside our range (the next one ending in $7$ would be $77$).
For $u = 9$, the multiples of $9$ are $9, 18, 27, 36, 45$ — again none ends in $9$ (the next would be $99$).
Both contribute zero.

$$u=7: \{\,\} \to 0. \quad u=9: \{\,\} \to 0.$$

💡 A multiple of $u$ that also ends in $u$ must come from $u \cdot k$ where $k \equiv 1 \pmod{10/\gcd(u,10)}$. For $u = 7, 9$ the next such $k$ after $1$ is $11$, giving $77, 99$ — both outside $[11, 49]$.

#2 Make a Systematic List 3.NBT.A.2 Step 5

Add the counts across all $9$ cases for the final total.

$$\underbrace{4}_{u=1} + \underbrace{4}_{u=2} + \underbrace{1}_{u=3} + \underbrace{2}_{u=4} + \underbrace{4}_{u=5} + \underbrace{1}_{u=6} + \underbrace{0}_{u=7} + \underbrace{1}_{u=8} + \underbrace{0}_{u=9} = 17 \;\Rightarrow\; \textbf{(C)}$$

💡 The systematic list is exhaustive and non-overlapping (every number has exactly one units digit), so adding the case counts gives the total without double-counting.

[1] #2 4.NBT.A.2 Set up the casework. Fix the units digit $u$ from $1$ to $9$ (skip $u = 0$ since

[2] #3 4.OA.B.4 Handle the easy units digits — the ones where the divisibility rule guarantees a

[3] #3 3.OA.C.7 Now the medium cases. For $u = 4$, check each candidate: $14, 24, 34, 44$. Divid

[4] #3 4.OA.B.4 The hard-but-empty cases. For $u = 7$, the multiples of $7$ are $7, 14, 21, 28,

[5] #2 3.NBT.A.2 Add the counts across all $9$ cases for the final total.

Review

Reasonableness: Write the $17$ winners out by tens digit to confirm none was missed or counted twice. Tens $1$: $11, 12, 15$ ($3$ numbers — $14$ fails because $14 / 4$ is not whole, $13, 16, 17, 18, 19$ also fail). Tens $2$: $21, 22, 24, 25$ ($4$). Tens $3$: $31, 32, 33, 35, 36$ ($5$). Tens $4$: $41, 42, 44, 45, 48$ ($5$). Total $3 + 4 + 5 + 5 = 17$ — matches the count by units digit. Order-of-magnitude check: each units digit averages roughly $17 / 9 \approx 1.9$ winners out of $4$ candidates, about $47\%$, which is exactly what you would expect when half the digits divide "easy" and half divide almost nothing.

Alternative: Tool #6 (Guess and Check) by direct brute force: just walk through $11, 12, 13, \dots, 49$ in order and check divisibility for each. That is $39$ checks instead of nine grouped cases — slower, more error-prone, and you get the same $17$. The case-by-units-digit list (Tool #2) is faster because it lets one divisibility rule clear $4$ candidates at once.

CCSS standards used (min grade 4)

4.OA.B.4 Find all factor pairs for a whole number in the range 1-100; determine whether a given whole number is a multiple of a given one-digit number (Deciding for each candidate $N \in \{11, \dots, 49\}$ whether $N$ is a multiple of its units digit $u$ — the exact "is $N$ a multiple of $u$?" task this standard names.)
4.NBT.A.2 Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form (Reading the units digit $u$ of each two-digit number $N$ (writing $N = 10 \cdot t + u$ with $t \in \{1,2,3,4\}$) so that the casework can be organized by $u$.)
3.OA.C.7 Fluently multiply and divide within 100 (Performing the actual divisibility checks ($14 \div 4$, $24 \div 4$, $33 \div 3$, $36 \div 6$, $48 \div 8$, etc.) using known multiplication facts.)
3.NBT.A.2 Fluently add and subtract within 1000 (Summing the nine case counts $4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ to get the final answer.)

⭐ Sort the $39$ numbers by their last digit and check each group with one divisibility rule. Three easy digits ($1, 2, 5$) give $12$ winners on their own, and the rest contribute $5$ more — total $17$, answer (C).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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