AMC 8 · 2005 · #18
Easy mode Grade 4Problem
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.
AMC 8 2005 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.
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 $13$.
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) $77$
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
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.
💡 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.
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 $8$.)
💡 Dividing $100$ by $13$ tells you roughly where the boundary sits; the integer just above the quotient is your first valid $k$.
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.)
💡 Same idea at the top end: divide $999$ by $13$ and round down. Two cheap multiplications confirm you have not slipped past the boundary.
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 $(\text{last} - \text{first}) + 1$.
💡 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.
4.OA.B.4 Rewrite the problem in terms of $k$. Every multiple of $13$ has the form $13k$, 4.NBT.B.6 Guess and check for the smallest $k$. Try $k = 7$: $13 \times 7 = 91$, only two 4.NBT.B.6 Guess and check for the largest $k$. Try $k = 77$: $13 \times 77 = 1001$, four d 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.4Find 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.6Find 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.8Solve 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 $(76 - 8) + 1 = 69$.
⭐ 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 $(76 - 8) + 1 = 69$.
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