AMC 10 · 2005 · #13
Grade 6 arithmeticPick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Count the multiples of 3
Multiples of 3 up to 2005: divide and keep the whole part — 2005 ÷ 3 leaves quotient 668, remainder 1.
Counting multiples of 3 up to a number is the same as asking how many times 3 fits inside it.
4.OA.B.4Identify SubproblemsCount the multiples of 4
Same move for multiples of 4: 2005 ÷ 4 leaves quotient 501, remainder 1.
The whole-number part of the division tells you how many full steps of 4 land at or below 2005.
4.NBT.B.6Identify SubproblemsThe overlap is the multiples of 12
A number sits in both circles only if lcm(3,4) = 12 divides it, so the overlap is exactly what we must exclude.
Being a multiple of both 3 and 4 at once is exactly being a multiple of their least common multiple.
Being a multiple of both three and four at once is exactly being a multiple of twelve.
▸ Why?
Meeting two divisibility rules at once is meeting the one rule for their least common multiple.
▸ Why?
Three and four share no prime, so their recipes simply combine with nothing counted twice.
Count the multiples of 12
Size that shared middle the same way: 2005 ÷ 12 leaves quotient 167, remainder 1.
The overlap of the two circles has its own tidy count: how many times 12 fits inside 2005.
6.NS.B.2Identify SubproblemsRemove the middle and add the crescents
Delete the middle from each circle: 668 - 167 = 501, 501 - 167 = 334; the crescents are disjoint, so 501 + 334 = 835 → (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.A.3Change Focus Count The ComplementDraw 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.
- Count the multiples of 3
- Count the multiples of 4
- The overlap is the multiples of 12
- Count the multiples of 12
- Remove the middle and add the crescents
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