AMC 10 · 2005 · #13
Grade 6 arithmeticHow many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12?
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: Look at the whole numbers from $1$ to $2005$. Count how many of them are a multiple of $3$ or a multiple of $4$, but throw out any that are a multiple of $12$.
Givens: The range of numbers is $1$ through $2005$; A number counts if it is a multiple of $3$ or a multiple of $4$ (or both); A number is excluded if it is a multiple of $12$; Answer choices: (A) $501$, (B) $668$, (C) $835$, (D) $1002$, (E) $1169$
Unknowns: How many numbers in the range are multiples of $3$ or $4$ but not $12$
Understand
Restated: Look at the whole numbers from $1$ to $2005$. Count how many of them are a multiple of $3$ or a multiple of $4$, but throw out any that are a multiple of $12$.
Givens: The range of numbers is $1$ through $2005$; A number counts if it is a multiple of $3$ or a multiple of $4$ (or both); A number is excluded if it is a multiple of $12$; Answer choices: (A) $501$, (B) $668$, (C) $835$, (D) $1002$, (E) $1169$
Plan
Primary tool: #12 Draw a Venn Diagram
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement
Tool #12 (Draw a Venn Diagram) fits "or but not" perfectly: draw one circle for multiples of $3$ and one for multiples of $4$. Where they overlap sits every number that is a multiple of both — which is exactly the multiples of $12$. The phrase "multiples of $3$ or $4$ but not $12$" is then just the two outer crescents, with the shared middle removed. Tool #7 (Identify Subproblems) supplies the raw counts for each circle and the overlap by simple division. Tool #16 (Change Focus / Count the Complement) is the "but not $12$" move: inside each circle we take away the shared middle, leaving only the part we actually want.
Execute — Answer: C
4.OA.B.4 Step 1 Count the multiples of 3
- The multiples of $3$ up to $2005$ are $3, 6, 9, \ldots$ up to the largest one that does not pass $2005$.
- How many are there?
- Divide and keep the whole-number part: $2005 \div 3 = 668$ remainder $1$, so there are $668$ multiples of $3$.
💡 Counting multiples of $3$ up to a number is the same as asking how many times $3$ fits inside it.
4.NBT.B.6 Step 2 Count the multiples of 4
- Do the same for multiples of $4$: $4, 8, 12, \ldots$ up to $2005$.
- Divide and keep the whole-number part: $2005 \div 4 = 501$ remainder $1$, so there are $501$ multiples of $4$.
💡 The whole-number part of the division tells you how many full steps of $4$ land at or below $2005$.
6.NS.B.4 Step 3 The overlap is the multiples of 12
- A number sits in both circles only if it is a multiple of $3$ and a multiple of $4$ at the same time.
- The smallest number that is a multiple of both is their least common multiple, $\text{lcm}(3,4) = 12$ (since $3$ and $4$ share no common factor, you just multiply them).
- So the overlap of the two circles is precisely the multiples of $12$ — the very numbers the problem tells us to exclude.
💡 Being a multiple of both $3$ and $4$ at once is exactly being a multiple of their least common multiple.
6.NS.B.2 Step 4 Count the multiples of 12
- Now count how many multiples of $12$ are in the range: $12, 24, 36, \ldots$ up to $2005$.
- Divide and keep the whole-number part: $2005 \div 12 = 167$ remainder $1$, so there are $167$ numbers in the shared middle.
💡 The overlap of the two circles has its own tidy count: how many times $12$ fits inside $2005$.
4.OA.A.3 Step 5 Remove the middle and add the crescents
- "But not $12$" means delete the shared middle from each circle.
- The multiples-of-$3$ circle keeps $668 - 167 = 501$ numbers; the multiples-of-$4$ circle keeps $501 - 167 = 334$ numbers.
- These two leftover crescents do not overlap anymore, so just add them: $501 + 334 = 835$.
- That is choice (C).
💡 Once the shared middle is gone from both circles, what is left are two separate pieces you can safely add without double-counting.
4.OA.B.4 The multiples of $3$ up to $2005$ are $3, 6, 9, \ldots$ up to the largest one th 4.NBT.B.6 Do the same for multiples of $4$: $4, 8, 12, \ldots$ up to $2005$. Divide and ke 6.NS.B.4 A number sits in both circles only if it is a multiple of $3$ and a multiple of 6.NS.B.2 Now count how many multiples of $12$ are in the range: $12, 24, 36, \ldots$ up t 4.OA.A.3 "But not $12$" means delete the shared middle from each circle. The multiples-of Review
Reasonableness: Cross-check with the union formula. The numbers that are a multiple of $3$ or $4$ (before the $12$ rule) total $668 + 501 - 167 = 1002$ — that is answer (D), the trap for anyone who forgets "but not $12$." Removing the $167$ multiples of $12$ from that union gives $1002 - 167 = 835$, matching. The other traps line up too: (E) $1169 = 668 + 501$ forgets the overlap completely, (B) $668$ is only the multiples of $3$, and (A) $501$ is only the multiples of $4$. Only (C) $835$ correctly keeps the two crescents.
Alternative: Count how many times each multiple of $12$ is over-counted. Adding the two circles blindly gives $668 + 501 = 1169$, and every multiple of $12$ was counted twice there (once as a multiple of $3$, once as a multiple of $4$). Since we want those numbers counted zero times, subtract them twice: $1169 - 2 \times 167 = 1169 - 334 = 835$. Same answer, reached by fixing the double-count directly instead of trimming each circle.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Recognizing and counting the multiples of $3$ up to $2005$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $2005$ by $4$ and keeping the whole-number part to count the multiples of $4$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Seeing that the overlap of multiples of $3$ and $4$ is the multiples of $\text{lcm}(3,4)=12$.)6.NS.B.2Fluently divide multi-digit numbers using the standard algorithm (Dividing $2005$ by the two-digit divisor $12$ to count the $167$ multiples of $12$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Subtracting the overlap from each circle and adding the two crescents to reach $835$.)
⭐ Draw two overlapping circles, notice the overlap is the multiples of the two numbers' least common multiple, then take that shared middle out before you add.
⭐ Draw two overlapping circles, notice the overlap is the multiples of the two numbers' least common multiple, then take that shared middle out before you add.
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