AMC 8 · 2018 · #21
Grade 6 number-theoryProblem
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
Pick an answer.
AMC 8 2018 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: We want to count the positive three-digit integers $N$ (so $100 \le N \le 999$) that leave a remainder of $2$ when divided by $6$, a remainder of $5$ when divided by $9$, and a remainder of $7$ when divided by $11$ — all three conditions at once.
Givens: $N \div 6$ leaves remainder $2$; $N \div 9$ leaves remainder $5$; $N \div 11$ leaves remainder $7$; $N$ is a three-digit positive integer ($100 \le N \le 999$); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The count of positive three-digit integers $N$ that satisfy all three remainder conditions
Understand
Restated: We want to count the positive three-digit integers $N$ (so $100 \le N \le 999$) that leave a remainder of $2$ when divided by $6$, a remainder of $5$ when divided by $9$, and a remainder of $7$ when divided by $11$ — all three conditions at once.
Givens: $N \div 6$ leaves remainder $2$; $N \div 9$ leaves remainder $5$; $N \div 11$ leaves remainder $7$; $N$ is a three-digit positive integer ($100 \le N \le 999$); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #2 Make a Systematic List
The three remainder conditions look messy until you compare each divisor with its remainder: $6 - 2 = 4$, $9 - 5 = 4$, $11 - 7 = 4$. That repeating $4$ is a Tool #5 pattern: in every case, $N$ is exactly $4$ short of a multiple of the divisor. So $N + 4$ is a common multiple of $6$, $9$, and $11$ — meaning $N + 4$ must be a multiple of $\operatorname{lcm}(6, 9, 11) = 198$. Once we have $N = 198k - 4$, Tool #2 (Systematic List) takes over: list the values for $k = 1, 2, 3, \ldots$ in order and stop as soon as we leave the three-digit range, then count.
Execute — Answer: E
4.NBT.B.6 Step 1 - Spot the pattern hidden in the three conditions.
- For each divisor $d$ and remainder $r$, compute $d - r$: $6 - 2 = 4$, $9 - 5 = 4$, $11 - 7 = 4$.
- The same gap of $4$ appears every time, which means $N$ is always $4$ less than a multiple of each divisor — equivalently, $N + 4$ is a multiple of $6$, of $9$, and of $11$ at the same time.
💡 Knowing that "divisor minus remainder" measures how far $N$ sits below the next multiple is a Grade 4 division-with-remainders idea.
6.NS.B.4 Step 2 - Find the smallest number that is a multiple of $6$, $9$, and $11$ all at once — that is the least common multiple.
- Factor each: $6 = 2 \times 3$, $9 = 3^{2}$, $11 = 11$.
- Take the highest power of each prime: $2 \times 3^{2} \times 11 = 198$.
- So $N + 4$ must be a multiple of $198$.
💡 Combining three divisibility requirements into one LCM is exactly the Grade 6 LCM standard at work.
4.OA.C.5 Step 3 - Write the general form.
- Since $N + 4$ is some positive multiple of $198$, write $N + 4 = 198k$ for a positive integer $k$, which gives $N = 198k - 4$.
- Every $N$ satisfying all three remainder conditions has this shape.
💡 Describing every solution with one rule $N = 198k - 4$ is a Grade 4 "number pattern from a rule" move.
4.NBT.B.4 Step 4 - Make a systematic list of $N = 198k - 4$ for $k = 1, 2, 3, \ldots$ and keep only the three-digit values.
- Stop as soon as $N$ would exceed $999$.
💡 Multiplying $198$ by small whole numbers and subtracting $4$ is Grade 4 multi-digit arithmetic — no tricks required.
4.OA.A.3 Step 5 - Count the valid $N$ values from the list.
- There are exactly five: $194, 392, 590, 788, 986$.
- So the answer is (E) $5$.
💡 Counting how many items in a list meet a condition is a Grade 4 multi-step word-problem skill.
4.NBT.B.6 Spot the pattern hidden in the three conditions. For each divisor $d$ and remain 6.NS.B.4 Find the smallest number that is a multiple of $6$, $9$, and $11$ all at once — 4.OA.C.5 Write the general form. Since $N + 4$ is some positive multiple of $198$, write 4.NBT.B.4 Make a systematic list of $N = 198k - 4$ for $k = 1, 2, 3, \ldots$ and keep only 4.OA.A.3 Count the valid $N$ values from the list. There are exactly five: $194, 392, 590 Review
Reasonableness: Spot-check the smallest candidate $N = 194$: $194 \div 6 = 32$ remainder $2$ ($\checkmark$), $194 \div 9 = 21$ remainder $5$ ($\checkmark$), $194 \div 11 = 17$ remainder $7$ ($\checkmark$). The solutions are spaced exactly $198$ apart, so the count is the number of multiples of $198$ in $[104, 1003]$, which is $\lfloor 1003/198 \rfloor - \lceil 104/198 \rceil + 1 = 5 - 1 + 1 = 5$. That matches the answer (E).
Alternative: Tool #6 (Guess and Check) on the multiples of $198$: list $198, 396, 594, 792, 990, 1188, \ldots$, subtract $4$ from each, and keep those in $[100, 999]$. You immediately get $194, 392, 590, 788, 986$ and stop — same count of $5$, no LCM theory needed if you trust the pattern.
CCSS standards used (min grade 6)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Reading each remainder condition as "$N$ is $r$ more than a multiple of $d$" and then noticing that $d - r = 4$ in all three cases — the divide-with-remainder picture is exactly the Grade 4 standard.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Computing $\operatorname{lcm}(6, 9, 11) = 198$ via prime factorization so we know how far apart consecutive solutions sit.)4.OA.C.5Generate a number or shape pattern following a given rule (Writing every solution with the single rule $N = 198k - 4$, $k = 1, 2, 3, \ldots$)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $198 \times k - 4$ for $k = 1, 2, 3, 4, 5, 6$ to list the candidate three-digit values.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Counting how many of the listed $N$ values fall in the three-digit range $[100, 999]$ to get the final answer.)
⭐ This AMC 8 problem only needs Grade 6 least common multiple plus a clever "divisor minus remainder" pattern you already know!
⭐ This AMC 8 problem only needs Grade 6 least common multiple plus a clever "divisor minus remainder" pattern you already know!
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