AMC 8 · 2017 · #12
Grade 4 number-theoryProblem
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?
Pick an answer.
AMC 8 2017 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 positive integer $N > 1$ such that dividing $N$ by $4$, by $5$, and by $6$ each leaves a remainder of $1$. Then say which of the given ranges contains $N$.
Givens: $N$ is a positive integer with $N > 1$; $N \div 4$ leaves remainder $1$; $N \div 5$ leaves remainder $1$; $N \div 6$ leaves remainder $1$; Answer choices: (A) $2$ and $19$, (B) $20$ and $39$, (C) $40$ and $59$, (D) $60$ and $79$, (E) $80$ and $124$
Unknowns: The smallest such $N$ (and the answer-choice range that contains it)
Understand
Restated: Find the smallest positive integer $N > 1$ such that dividing $N$ by $4$, by $5$, and by $6$ each leaves a remainder of $1$. Then say which of the given ranges contains $N$.
Givens: $N$ is a positive integer with $N > 1$; $N \div 4$ leaves remainder $1$; $N \div 5$ leaves remainder $1$; $N \div 6$ leaves remainder $1$; Answer choices: (A) $2$ and $19$, (B) $20$ and $39$, (C) $40$ and $59$, (D) $60$ and $79$, (E) $80$ and $124$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
Three simultaneous remainder conditions feel intimidating, but Tool #9 transforms the problem into something much easier: "$N$ has remainder $1$ when divided by $d$" is the same as saying "$N - 1$ is a multiple of $d$." So instead of hunting for $N$ directly, we hunt for $N - 1$, which must be a common multiple of $4$, $5$, and $6$ — pure 4th-grade multiples thinking. Tool #2 (Systematic List) then sweeps multiples of the largest divisor ($6$) in order and checks each against the other two conditions, which is faster than computing a formal LCM. Tool #3 (Eliminate) closes the problem by matching $N$ against the five labeled ranges.
Execute — Answer: D
4.NBT.B.6 Step 1 - Rewrite the three remainder conditions as one divisibility condition.
- "$N$ leaves remainder $1$ when divided by $d$" means $N - 1$ is exactly divisible by $d$.
- Applying this to $d = 4, 5, 6$ turns the problem into: find the smallest positive $N - 1$ that is a common multiple of $4$, $5$, and $6$.
💡 Remainder $1$ means "one past a multiple," so subtracting $1$ snaps the number back onto a multiple — a Grade 4 remainder idea.
4.OA.B.4 Step 2 - List multiples of the largest divisor ($6$) in order and circle the ones that are also multiples of both $4$ and $5$.
- Going through $6, 12, 18, 24, 30, 36, 42, 48, 54, 60$, only $60$ is divisible by $4$ (since $60 = 4 \times 15$) and by $5$ (since $60 = 5 \times 12$).
- So the smallest positive common multiple of $4$, $5$, $6$ is $60$.
💡 Sweeping multiples of one number and checking divisibility by the others is the Grade 4 "factors and multiples" skill in action.
4.OA.A.3 Step 3 - Add $1$ to recover $N$.
- Since the smallest valid value of $N - 1$ is $60$, the smallest valid $N$ is $60 + 1 = 61$.
- (The next candidate would come from the next common multiple, $120$, giving $N = 121$ — much larger, so $61$ wins.)
💡 Undoing the "$-1$" shift to translate between the easier problem and the original is exactly the Grade 4 multi-step word-problem move.
4.NBT.A.2 Step 4 - Match $N = 61$ against the five range choices and eliminate the rest.
- $61$ is not in $[2, 19]$, $[20, 39]$, or $[40, 59]$, and it is below $80$ so not in $[80, 124]$.
- It fits cleanly into $[60, 79]$, which is choice (D).
💡 Comparing $61$ to the range endpoints uses Grade 4 multi-digit number comparison.
4.NBT.B.6 Rewrite the three remainder conditions as one divisibility condition. "$N$ leave 4.OA.B.4 List multiples of the largest divisor ($6$) in order and circle the ones that ar 4.OA.A.3 Add $1$ to recover $N$. Since the smallest valid value of $N - 1$ is $60$, the s 4.NBT.A.2 Match $N = 61$ against the five range choices and eliminate the rest. $61$ is no Review
Reasonableness: Quick check: $61 \div 4 = 15$ remainder $1$, $61 \div 5 = 12$ remainder $1$, $61 \div 6 = 10$ remainder $1$. All three remainder conditions hold. And $61$ is greater than $1$, so the "greater than 1" clause is satisfied. The only smaller candidate would come from a common multiple of $4$, $5$, $6$ that is smaller than $60$ — but our systematic list confirmed none exist. So $N = 61$ is genuinely the smallest, and it sits in the $60$–$79$ range, matching (D).
Alternative: Tool #6 (Guess and Check) directly on the answer choices: pick the smallest $N$ in each range and test the three remainders. (A) $N = 5$: $5 \div 4$ has remainder $1$, but $5 \div 5 = 1$ remainder $0$ — fail. (B) $N = 21$: $21 \div 4$ has remainder $1$, $21 \div 5$ has remainder $1$, but $21 \div 6 = 3$ remainder $3$ — fail. (C) $N = 41$: $41 \div 4$ has remainder $1$, $41 \div 5$ has remainder $1$, but $41 \div 6 = 6$ remainder $5$ — fail. (D) $N = 61$: all three give remainder $1$ — works. So (D).
CCSS standards used (min grade 4)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Reading the phrase "leaves a remainder of $1$" as the divisibility statement "$N - 1$ is exactly divisible by $d$" — the Grade 4 remainder concept made explicit.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing multiples of $6$ in order and checking each one for divisibility by $4$ and $5$ until the first common multiple ($60$) appears.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Translating between the helper quantity $N - 1$ and the original $N$ by adding $1$ at the end — the multi-step bookkeeping of the word problem.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Comparing $N = 61$ against the endpoints of the five labeled ranges ($19, 39, 59, 79, 124$) to choose the correct interval.)
⭐ This AMC 8 problem only needs Grade 4 multiples and remainders you already know — "remainder 1" just means "one more than a multiple"!
⭐ This AMC 8 problem only needs Grade 4 multiples and remainders you already know — "remainder 1" just means "one more than a multiple"!
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