AMC 10 · 2005 · #22
Grade 6 arithmeticLet S be the set of the 2005 smallest positive multiples of 4, and let T be the set of the 2005 smallest positive multiples of 6. How many elements are common to S and T?
Pick an answer.
AMC 10 2005 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: $S$ is the list of the $2005$ smallest positive multiples of $4$, and $T$ is the list of the $2005$ smallest positive multiples of $6$. Count how many numbers appear in both lists.
Givens: $S = \{4, 8, 12, \dots\}$ — the first $2005$ multiples of $4$; $T = \{6, 12, 18, \dots\}$ — the first $2005$ multiples of $6$; Each list has exactly $2005$ numbers; Answer choices: (A) $166$, (B) $333$, (C) $500$, (D) $668$, (E) $1001$
Unknowns: How many numbers belong to both $S$ and $T$
Understand
Restated: $S$ is the list of the $2005$ smallest positive multiples of $4$, and $T$ is the list of the $2005$ smallest positive multiples of $6$. Count how many numbers appear in both lists.
Givens: $S = \{4, 8, 12, \dots\}$ — the first $2005$ multiples of $4$; $T = \{6, 12, 18, \dots\}$ — the first $2005$ multiples of $6$; Each list has exactly $2005$ numbers; Answer choices: (A) $166$, (B) $333$, (C) $500$, (D) $668$, (E) $1001$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #14 Extreme Principle
A number shared by both lists must be a multiple of $4$ and of $6$ at the same time, so it must be a multiple of their least common multiple, $12$. Tool #4 names any such shared number $12k$, turning "how many common elements" into "how many whole numbers $k$ work." Tool #7 splits the job into two smaller questions — find the biggest number each list reaches — and Tool #14 picks the tighter of the two ceilings, which is the only one that actually limits the count. Then counting the valid $k$ is one division.
Execute — Answer: D
4.OA.B.4 Step 1 How far each list reaches
- Find the largest number in each list.
- The last (biggest) multiple of $4$ in $S$ is the $2005$th one, and the last multiple of $6$ in $T$ is the $2005$th one.
💡 The $2005$th multiple of a number is just that number times $2005$.
6.NS.B.4 Step 2 A shared number is a multiple of 12
- Any number in both lists is a multiple of $4$ and a multiple of $6$, so it is a multiple of the least common multiple of $4$ and $6$, which is $12$.
- Name such a shared number $12k$ for some whole number $k \ge 1$.
💡 To sit in both lists a number must be built from $4$s and from $6$s at once, and $12$ is the smallest number that is both.
6.EE.B.8 Step 3 Which ceiling actually limits it
- For $12k$ to really appear in $S$ it must not exceed $S$'s largest element, and likewise for $T$.
- Both conditions must hold, so the smaller ceiling is the one that binds.
- Since $8020 < 12030$, the deciding requirement is $12k \le 8020$.
💡 $S$ stops sooner than $T$, so a shared number can only go as high as $S$ allows.
6.NS.B.2 Step 4 Count the valid multiples
- Count the whole numbers $k \ge 1$ with $12k \le 8020$.
- Divide $8020$ by $12$ and keep the whole-number part, since $k$ must be a whole number.
💡 Each whole number $k$ from $1$ up to the cutoff gives exactly one shared number $12k$.
4.OA.B.4 Find the largest number in each list. The last (biggest) multiple of $4$ in $S$ 6.NS.B.4 Any number in both lists is a multiple of $4$ and a multiple of $6$, so it is a 6.EE.B.8 For $12k$ to really appear in $S$ it must not exceed $S$'s largest element, and 6.NS.B.2 Count the whole numbers $k \ge 1$ with $12k \le 8020$. Divide $8020$ by $12$ and Review
Reasonableness: There are $668$ shared numbers, choice (D). A quick sanity check: the multiples of $12$ inside $S$ are every $3$rd element of $S$ (since $12 = 4 \times 3$), and $2005 \div 3 \approx 668$, matching. All of these also land in $T$ because $T$ reaches even higher ($12030 > 8020$). The tempting trap $1001$ (E) is roughly $2005/2$ — the multiples of $12$ inside $T$ — but many of those sit above $8020$ and never appear in $S$, so they should not be counted.
Alternative: List the shared multiples of $12$ directly. Inside $S$ they run $12, 24, \dots$ up to $8020$, so the last one is $12 \times \lfloor 8020/12 \rfloor = 12 \times 668 = 8016$. The shared numbers are $12, 24, \dots, 8016$, and counting them gives $8016 \div 12 = 668$. Same answer.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing the $2005$th multiple of $4$ and of $6$ to find how far each list reaches.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Using $\operatorname{lcm}(4,6) = 12$ to show every shared number is a multiple of $12$.)6.EE.B.8Write an inequality of the form x > c or x < c and graph on a number line (Writing the range conditions $12k \le 8020$ and $12k \le 12030$ and keeping the tighter one.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Dividing $8020$ by $12$ to count how many multiples of $12$ fit.)
⭐ Numbers in both lists must be multiples of the LCM ($12$), so count the multiples of $12$ that fit under the shorter list's ceiling.
⭐ Numbers in both lists must be multiples of the LCM ($12$), so count the multiples of $12$ that fit under the shorter list's ceiling.
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