AMC 8 · 2005 · #18

Easy mode Grade 4

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Problem

A three-digit number is any whole number from $100$ to $999$ .

How many of these three-digit numbers can be divided exactly by $13$ (with no remainder)?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Count how many three-digit numbers (the integers from $100$ to $999$) are divisible by $13$.

Givens: A three-digit number is an integer from $100$ to $999$, inclusive; A number is divisible by $13$ when it equals $13k$ for some positive integer $k$; Answer choices: (A) $7$, (B) $67$, (C) $69$, (D) $76$, (E) $77$

Unknowns: The number of three-digit multiples of $13$

Understand

Restated: Count how many three-digit numbers (the integers from $100$ to $999$) are divisible by $13$.

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #6 Guess and Check

Counting three-digit multiples of $13$ directly is awkward — the multiples ($104, 117, 130, \dots$) are not the numbers we want to count. Tool #9 (Easier Related Problem) replaces the task with something much simpler: each three-digit multiple has the form $13k$, so we only need to count the values of $k$ that make $13k$ a three-digit number. That turns the question into "how many integers $k$ satisfy $100 \le 13k \le 999$?" — a plain integer-counting problem. Tool #6 (Guess and Check) is the right way to find the two endpoints of $k$: try a few values near $100/13$ and $999/13$ until $13k$ first lands in, and last lands inside, the three-digit window. Once the endpoints are pinned down, the count is just $(\text{last} - \text{first}) + 1$.

Execute — Answer: C

#9 Solve an Easier Related Problem 4.OA.B.4 Step 1

Rewrite the problem in terms of $k$.
Every multiple of $13$ has the form $13k$, so a three-digit multiple of $13$ corresponds to an integer $k$ with $100 \le 13k \le 999$.
Counting three-digit multiples and counting these $k$ values give the same answer.

$$100 \le 13k \le 999,\ \ k \in \{1, 2, 3, \dots\}$$

💡 Every multiple of $13$ is $13 \times \text{something}$. So instead of hunting for three-digit multiples, hunt for the multipliers $k$ — they form a tidy block of consecutive integers.

#6 Guess and Check 4.NBT.B.6 Step 2

Guess and check for the smallest $k$.
Try $k = 7$: $13 \times 7 = 91$, only two digits.
Try $k = 8$: $13 \times 8 = 104$, three digits.
So the smallest qualifying $k$ is $8$.
(You can also see this from $100 \div 13 \approx 7.69$, so the first integer above $7.69$ is $8$.)

$$13 \times 7 = 91 < 100,\ \ 13 \times 8 = 104 \ge 100 \;\Rightarrow\; k_{\min} = 8$$

💡 Dividing $100$ by $13$ tells you roughly where the boundary sits; the integer just above the quotient is your first valid $k$.

#6 Guess and Check 4.NBT.B.6 Step 3

Guess and check for the largest $k$.
Try $k = 77$: $13 \times 77 = 1001$, four digits — too big.
Try $k = 76$: $13 \times 76 = 988$, three digits — good.
So the largest qualifying $k$ is $76$.
(Equivalently, $999 \div 13 \approx 76.85$, so take the integer just below.)

$$13 \times 77 = 1001 > 999,\ \ 13 \times 76 = 988 \le 999 \;\Rightarrow\; k_{\max} = 76$$

💡 Same idea at the top end: divide $999$ by $13$ and round down. Two cheap multiplications confirm you have not slipped past the boundary.

#9 Solve an Easier Related Problem 3.OA.D.8 Step 4

Count the integers from $8$ to $76$ inclusive.
The valid values of $k$ are $8, 9, 10, \dots, 76$ — a block of consecutive integers — so the count is $(\text{last} - \text{first}) + 1$.

$$(76 - 8) + 1 = 68 + 1 = 69 \;\Rightarrow\; \textbf{(C)}$$

💡 When you count fenceposts from $8$ to $76$, you get one more than the difference. The $+1$ catches the first post you would otherwise miss.

[1] #9 4.OA.B.4 Rewrite the problem in terms of $k$. Every multiple of $13$ has the form $13k$,

[2] #6 4.NBT.B.6 Guess and check for the smallest $k$. Try $k = 7$: $13 \times 7 = 91$, only two

[3] #6 4.NBT.B.6 Guess and check for the largest $k$. Try $k = 77$: $13 \times 77 = 1001$, four d

[4] #9 3.OA.D.8 Count the integers from $8$ to $76$ inclusive. The valid values of $k$ are $8, 9

Review

Reasonableness: A quick estimate confirms the count. There are $900$ three-digit numbers ($100$ through $999$), and roughly $1$ in every $13$ of them is a multiple of $13$. That gives $900 / 13 \approx 69.2$, which matches the exact count of $69$. The answer choices $(A)\ 7$ and $(D)\ 76$ are red herrings — $7$ is the largest $k$ that is too small, and $76$ is the largest valid $k$, not the count itself. Choices $(B)\ 67$ and $(E)\ 77$ are off-by-one or off-by-two mistakes from the $(\text{last} - \text{first}) + 1$ step. Only $(C)\ 69$ survives.

Alternative: Tool #5 (Find a Pattern): in the integers from $1$ to $99$ there are $\lfloor 99/13 \rfloor = 7$ multiples of $13$, and from $1$ to $999$ there are $\lfloor 999/13 \rfloor = 76$. Subtracting removes the multiples of $13$ below $100$: $76 - 7 = 69$. Same answer, but expressed as "multiples up to $999$ minus multiples up to $99$".

CCSS standards used (min grade 4)

4.OA.B.4 Find all factor pairs for a whole number; recognize that a whole number is a multiple of each of its factors (Recognizing that every multiple of $13$ has the form $13k$ and rewriting the question as a count of valid $k$ values.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors (Dividing $100$ and $999$ by $13$ (and checking with multiplication) to locate the smallest and largest $k$ inside the three-digit window.)
3.OA.D.8 Solve two-step word problems using the four operations (Counting consecutive integers from $8$ to $76$ with the $(\text{last} - \text{first}) + 1$ formula to reach the final answer $69$.)

⭐ Instead of counting three-digit multiples of $13$, count the multipliers $k$. The smallest is $8$ (since $13 \times 8 = 104$), the largest is $76$ (since $13 \times 76 = 988$), and the integers from $8$ to $76$ number $(76 - 8) + 1 = 69$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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