AMC 8 · 2006 · #23
Grade 6 number-theoryProblem
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
Pick an answer.
AMC 8 2006 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 whole number $N$ that leaves remainder $4$ when divided by $6$ and remainder $3$ when divided by $5$. Then report the remainder when that $N$ is divided by $7$.
Givens: $N$ divided by $6$ leaves remainder $4$; $N$ divided by $5$ leaves remainder $3$; $N$ is the smallest positive whole number meeting both conditions; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $5$
Unknowns: The smallest $N$ satisfying both remainder conditions; The remainder when $N$ is divided by $7$
Understand
Restated: Find the smallest positive whole number $N$ that leaves remainder $4$ when divided by $6$ and remainder $3$ when divided by $5$. Then report the remainder when that $N$ is divided by $7$.
Givens: $N$ divided by $6$ leaves remainder $4$; $N$ divided by $5$ leaves remainder $3$; $N$ is the smallest positive whole number meeting both conditions; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $5$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #5 Look for a Pattern
Each remainder condition pins $N$ to a clear arithmetic sequence — "start at $4$, step by $6$" for the first, "start at $3$, step by $5$" for the second — so Tool #2 (Systematic List) writes both sequences cleanly and the first shared value is the smallest $N$. Tool #5 (Look for a Pattern) sharpens the search: the gap to the next multiple of $6$ is $2$, and the gap to the next multiple of $5$ is also $2$, so $N + 2$ must be a common multiple of $5$ and $6$. That pattern means we only need the smallest common multiple of $5$ and $6$, which is $30$, and then $N = 30 - 2 = 28$. Both tools land on the same $N$, after which the final remainder by $7$ is one division.
Execute — Answer: A
4.OA.C.5 Step 1 - Write the systematic list for the first condition.
- The numbers that leave remainder $4$ when divided by $6$ start at $4$ and go up by $6$ each time.
💡 Grade 4 "generate a number pattern from a rule" — the rule here is +6, starting at $4$.
4.OA.C.5 Step 2 - Write the systematic list for the second condition.
- The numbers that leave remainder $3$ when divided by $5$ start at $3$ and go up by $5$ each time.
💡 Same Grade 4 pattern move with a different rule: +5, starting at $3$.
4.OA.B.4 Step 3 - Scan both lists for the smallest shared value.
- Compare items in order; the first match is $28$, so $N = 28$ is the smallest coin count satisfying both conditions.
💡 Intersecting two short lists is the cleanest Grade 4 way to pin down the smallest number that fits two multiple-style rules at once.
4.NBT.B.6 Step 4 - Divide $N = 28$ by $7$ to read off the final remainder.
- Since $28 = 7 \times 4$, the remainder is $0$.
💡 Grade 4 division-with-remainder closes the problem: $7$ divides $28$ evenly, so nothing is left over.
4.OA.C.5 Write the systematic list for the first condition. The numbers that leave remain 4.OA.C.5 Write the systematic list for the second condition. The numbers that leave remai 4.OA.B.4 Scan both lists for the smallest shared value. Compare items in order; the first 4.NBT.B.6 Divide $N = 28$ by $7$ to read off the final remainder. Since $28 = 7 \times 4$, Review
Reasonableness: Verify both original conditions on $N = 28$. Dividing by $6$: $28 = 6 \times 4 + 4$, remainder $4$ — matches. Dividing by $5$: $28 = 5 \times 5 + 3$, remainder $3$ — matches. So $28$ really is a valid coin count, and no smaller positive integer appears in both lists, so it is the smallest. Dividing by $7$ gives $28 = 7 \times 4$, remainder $0$, so answer (A) is consistent.
Alternative: Tool #5 (Look for a Pattern) gives a shortcut. Notice that $4$ is $2$ short of $6$ and $3$ is $2$ short of $5$, so if there were $2$ more coins, the total would be divisible by both $5$ and $6$. The smallest common multiple of $5$ and $6$ is $30$, so $N + 2 = 30$, which forces $N = 28$. Then $28 \div 7$ leaves remainder $0$, again (A).
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Writing the two arithmetic sequences ($4, 10, 16, \ldots$ and $3, 8, 13, \ldots$) from the remainder conditions.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing the lists as multiples of $6$ shifted by $4$ and multiples of $5$ shifted by $3$, and finding the first common value $28$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing $28 \div 7 = 4$ remainder $0$, and verifying $28 \div 6$ and $28 \div 5$ in the reasonableness check.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Powering the shortcut: the least common multiple of $5$ and $6$ is $30$, so $N + 2 = 30$ and $N = 28$.)
⭐ When a count leaves two awkward remainders, two short arithmetic lists usually meet within a few terms — and spotting that both shortfalls are the same number turns the search into a quick LCM.
⭐ When a count leaves two awkward remainders, two short arithmetic lists usually meet within a few terms — and spotting that both shortfalls are the same number turns the search into a quick LCM.
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