AMC 8 · 2017 · #12
Easy mode Grade 4Problem
Imagine searching for a whole number bigger than 1 that does the following:
- When you divide it by 4, the remainder is 1.
- When you divide it by 5, the remainder is 1.
- When you divide it by 6, the remainder is 1.
Many numbers fit. We want the smallest one.
Find that smallest number. Which pair of numbers below does it sit between?
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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