AMC 8 · 2004 · #19
Grade 6 number-theoryProblem
A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3,4,5, and 6. The smallest such number lies between which two numbers?
Pick an answer.
AMC 8 2004 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 $N > 2$ such that dividing $N$ by each of $3, 4, 5,$ and $6$ leaves a remainder of $2$. Then identify the interval from the answer choices that contains $N$.
Givens: $N$ is a whole number with $N > 2$; $N \div 3, \; N \div 4, \; N \div 5, \; N \div 6$ each leave remainder $2$; $N$ is the smallest such number; Answer choices: (A) $40$ and $49$, (B) $60$ and $79$, (C) $100$ and $129$, (D) $210$ and $249$, (E) $320$ and $369$
Unknowns: The smallest qualifying $N$, and the interval that contains it
Understand
Restated: Find the smallest whole number $N > 2$ such that dividing $N$ by each of $3, 4, 5,$ and $6$ leaves a remainder of $2$. Then identify the interval from the answer choices that contains $N$.
Givens: $N$ is a whole number with $N > 2$; $N \div 3, \; N \div 4, \; N \div 5, \; N \div 6$ each leave remainder $2$; $N$ is the smallest such number; Answer choices: (A) $40$ and $49$, (B) $60$ and $79$, (C) $100$ and $129$, (D) $210$ and $249$, (E) $320$ and $369$
Plan
Primary tool: #7 Break into Subproblems
Secondary: #5 Find a Pattern
Four remainder conditions look like four separate problems, but Tool #5 (Find a Pattern) shows they all share the same shape: $N - 2$ is a multiple of $3$, of $4$, of $5$, and of $6$. That reframing collapses the four conditions into one: $N - 2$ is a common multiple of $\{3, 4, 5, 6\}$. Tool #7 (Break into Subproblems) then splits the work cleanly — first find the smallest common multiple (the LCM of the four divisors), then add the $+2$ offset to recover $N$. Two short steps replace a messy hunt.
Execute — Answer: B
4.OA.B.4 Step 1 - Spot the shared pattern.
- "$N$ leaves remainder $2$ when divided by $d$" means $N - 2$ is an exact multiple of $d$.
- Apply that rewrite to all four divisors at once.
💡 Grade 4 "multiples" language: a remainder of $2$ is just the multiple $N - 2$ shifted up by $2$.
6.NS.B.4 Step 2 - Subproblem 1: find the smallest positive value of $N - 2$.
- Since $N - 2$ must be a multiple of all four divisors, it is a common multiple of $\{3, 4, 5, 6\}$, and the smallest one is their LCM.
💡 Tool #7 splits the work: pin down $N - 2$ first, then handle the $+2$. Grade 6 number theory says the smallest shared multiple is the LCM.
6.NS.B.4 Step 3 - Compute the LCM from prime factorizations.
- Write each divisor as a product of primes, then take the highest power of each prime that appears.
💡 Highest power of $2$ is $2^2$ (from $4$); highest power of $3$ is $3^1$ (from $3$ or $6$); highest power of $5$ is $5^1$ (from $5$). Multiply: $4 \cdot 3 \cdot 5 = 60$.
4.OA.A.3 Step 4 - Subproblem 2: undo the shift.
- Recover $N$ by adding $2$ back to the LCM, then check that $N > 2$.
💡 $62 > 2$ holds, and $62$ falls inside the interval $60$-$79$ from choice (B).
4.OA.B.4 Spot the shared pattern. "$N$ leaves remainder $2$ when divided by $d$" means $N 6.NS.B.4 Subproblem 1: find the smallest positive value of $N - 2$. Since $N - 2$ must be 6.NS.B.4 Compute the LCM from prime factorizations. Write each divisor as a product of pr 4.OA.A.3 Subproblem 2: undo the shift. Recover $N$ by adding $2$ back to the LCM, then ch Review
Reasonableness: Verify each remainder directly: $62 = 3 \cdot 20 + 2$, $62 = 4 \cdot 15 + 2$, $62 = 5 \cdot 12 + 2$, $62 = 6 \cdot 10 + 2$. All four give remainder $2$, and $62 > 2$. The next qualifying number would be $60 + 60 + 2 = 122$, which lies in choice (C), so $62$ really is the smallest. Choice (A) $40$-$49$ is below the LCM and cannot work; choices (D) and (E) are far too large.
Alternative: Tool #2 (Make a Systematic List): the divisor $5$ forces $N$ to end in $2$ or $7$ (since $N - 2$ is a multiple of $5$). List candidates $7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, \ldots$ and test divisibility of $N - 2$ by $4$ (needs $N - 2$ multiple of $4$, so $N$ ends in $2$ with $N - 2$ a multiple of $4$): that narrows to $N = 22, 42, 62, 82, \ldots$. Now check divisibility of $N - 2$ by $3$: $22 - 2 = 20$ no; $42 - 2 = 40$ no; $62 - 2 = 60$ yes. (And $6 \mid 60$ is automatic once $3 \mid 60$ and $2 \mid 60$.) So $N = 62$, lying in interval (B).
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Rewriting each remainder condition as a multiple condition: $N - 2$ is a multiple of $3, 4, 5, 6$.)6.NS.B.4Find the greatest common factor and least common multiple of two whole numbers (Computing $\operatorname{lcm}(3, 4, 5, 6) = 60$ from prime factorizations $3, 2^2, 5, 2 \cdot 3$ by taking the highest power of each prime.)4.OA.A.3Solve multi-step word problems using the four operations with whole numbers (Adding the remainder offset $N = 60 + 2 = 62$ and confirming $62 > 2$ falls inside the interval $60 < 62 < 79$.)
⭐ When the same remainder shows up for several divisors, subtract that remainder first — what is left is a plain LCM, and adding the remainder back gives the answer.
⭐ When the same remainder shows up for several divisors, subtract that remainder first — what is left is a plain LCM, and adding the remainder back gives the answer.
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