AMC 8 · 2012 · #15
Grade 6 number-theoryProblem
The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?
Pick an answer.
AMC 8 2012 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: Find the smallest whole number bigger than $2$ that leaves a remainder of $2$ when you divide it by $3$, by $4$, by $5$, and by $6$. Then say which of the given ranges it falls inside.
Givens: The number $N$ we want is greater than $2$; $N$ divided by $3$ leaves remainder $2$; $N$ divided by $4$ leaves remainder $2$; $N$ divided by $5$ leaves remainder $2$; $N$ divided by $6$ leaves remainder $2$; Answer choices (ranges): (A) $40$–$50$, (B) $51$–$55$, (C) $56$–$60$, (D) $61$–$65$, (E) $66$–$99$
Unknowns: The smallest such $N$; Which answer-choice range contains that $N$
Understand
Restated: Find the smallest whole number bigger than $2$ that leaves a remainder of $2$ when you divide it by $3$, by $4$, by $5$, and by $6$. Then say which of the given ranges it falls inside.
Givens: The number $N$ we want is greater than $2$; $N$ divided by $3$ leaves remainder $2$; $N$ divided by $4$ leaves remainder $2$; $N$ divided by $5$ leaves remainder $2$; $N$ divided by $6$ leaves remainder $2$; Answer choices (ranges): (A) $40$–$50$, (B) $51$–$55$, (C) $56$–$60$, (D) $61$–$65$, (E) $66$–$99$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
The problem packs four divisibility conditions into one question. Tool #7 (Identify Subproblems) turns that into one clean step: subtract the common remainder of $2$, and the question becomes "what is the smallest positive number that is divisible by $3$, $4$, $5$, and $6$ at the same time?" — the definition of the least common multiple. Tool #5 (Look for a Pattern) handles the leftover comparison: every working $N$ has the form $60k + 2$, so we just pick the smallest $k$ and walk the pattern. Tool #3 (Eliminate Possibilities) then maps the resulting $N$ against the five range choices and rules out the four that miss. We avoid tool #13 (Algebra) and modular-arithmetic notation because the "shift by $2$" subproblem move makes them unnecessary.
Execute — Answer: D
4.OA.B.4 Step 1 - Pull the constant remainder out.
- Saying "$N$ leaves remainder $2$ when divided by $d$" is the same as saying "$N - 2$ is a multiple of $d$".
- Doing this for all four divisors at once turns the original problem into a much simpler one: find the smallest positive value of $N - 2$ that is a multiple of $3$, $4$, $5$, and $6$.
💡 Stripping the shared remainder is the Tool #7 subproblem move — one harder question becomes one easier question about plain multiples.
6.NS.B.4 Step 2 - Find the smallest positive number that is a multiple of $3$, $4$, $5$, and $6$ — the least common multiple.
- Use prime factorization: $3 = 3$, $4 = 2^2$, $5 = 5$, $6 = 2 \times 3$.
- Take the highest power of each prime that shows up.
💡 The $6$ is "free" because $6 = 2 \times 3$ is already covered by the $4$ and $3$ — no new prime is added.
4.OA.C.5 Step 3 - List the candidate values of $N$ in order and pick the first one that is greater than $2$.
- Since $N - 2$ must be a positive multiple of $60$, write $N - 2 = 60k$ for $k = 1, 2, 3, \dots$ and read off the pattern.
💡 Listing the first few terms of the "add $60$" pattern makes the smallest one — and the gap to the next one — obvious.
1.NBT.B.3 Step 4 - The smallest valid $N$ is $62$.
- Check the ranges in the answer choices: $40$–$50$, $51$–$55$, $56$–$60$, $61$–$65$, $66$–$99$.
- Only one of them contains $62$.
💡 Comparing $62$ against each range is just the Tool #3 "eliminate possibilities" move — four ranges miss, one fits.
4.OA.B.4 Pull the constant remainder out. Saying "$N$ leaves remainder $2$ when divided b 6.NS.B.4 Find the smallest positive number that is a multiple of $3$, $4$, $5$, and $6$ — 4.OA.C.5 List the candidate values of $N$ in order and pick the first one that is greater 1.NBT.B.3 The smallest valid $N$ is $62$. Check the ranges in the answer choices: $40$–$50 Review
Reasonableness: Verify $N = 62$ directly: $62 = 3 \times 20 + 2$ (remainder $2$, good), $62 = 4 \times 15 + 2$ (good), $62 = 5 \times 12 + 2$ (good), $62 = 6 \times 10 + 2$ (good). All four conditions hold. And nothing smaller works, because any smaller candidate $N - 2$ would have to be a positive multiple of $60$ smaller than $60$ — impossible. So $62$ is the smallest, and $62$ sits in the $61$–$65$ range, confirming (D).
Alternative: Tool #3 (Eliminate Possibilities) directly on the choices: any valid $N$ has the form $60k + 2$, so the only options inside the listed ranges are $62$ (range D) and $122$ (range E, but $122 > 99$, so E is also out for the smallest one). Range A ($40$–$50$), B ($51$–$55$), and C ($56$–$60$) cannot contain any number of the form $60k + 2$ at all, since $60(0) + 2 = 2$ is too small and $60(1) + 2 = 62$ is already past them. Only (D) survives.
CCSS standards used (min grade 6)
1.NBT.B.3Compare two two-digit numbers using <, =, > symbols (Checking that $62$ lies between $61$ and $65$ to pick the correct answer-choice range.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that "leaves remainder $2$ when divided by $d$" is the same as "$N - 2$ is a multiple of $d$", and stating this for all four divisors at once.)4.OA.C.5Generate a number or shape pattern following a given rule (Generating the list of valid $N$ values $62, 122, 182, \dots$ by repeatedly adding $60$ to $2$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Computing $\mathrm{LCM}(3, 4, 5, 6) = 60$ via prime factorization — the key number that drives the whole solution.)
⭐ This AMC 8 problem boils down to one Grade 6 idea — finding the least common multiple of $3, 4, 5, 6$ — once you spot that subtracting the shared remainder of $2$ makes the four conditions collapse into one!
⭐ This AMC 8 problem boils down to one Grade 6 idea — finding the least common multiple of $3, 4, 5, 6$ — once you spot that subtracting the shared remainder of $2$ makes the four conditions collapse into one!
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