AMC 8 · 2012 · #15
Easy mode Grade 6Problem
Find a whole number that is greater than 2 and has this special property: when you divide it by 3, the remainder is 2. When you divide it by 4, the remainder is 2. When you divide it by 5, the remainder is 2. When you divide it by 6, the remainder is also 2.
There are many such numbers. We want the smallest one.
Which of these ranges does that smallest number fall into?
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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