AMC 8 · 2000 · #11

Easy mode Grade 4

Problem

Look at the number $64$ . Its last digit (the ones place) is $4$ . Notice that $64$ can be divided evenly by $4$ — no remainder. So we say $64$ has a special property: it is divisible by its own last digit.

Now think about all the whole numbers between $10$ and $50$ .

How many of them have the same property — divisible by their own last digit?

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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