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2005 AMC 10B Problem 13

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2005 AMC 10B Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Inclusion-Exclusion · difficulty 2 of 5 · about 3 min · position P11-15

How many numbers between $1$ and $2005$ are integer multiples of $3$ or $4$ but not $12$ ?

A)
$501$
B)
$668$
C)
$835$
D)
$1002$
E)
$1169$

Answer

Key insight

Multiples of 12 are counted once in the 3-list and once in the 4-list, so to exclude them entirely subtract the 12-count twice.

Solution

Count each kind of multiple up to $2005$ :
$\left\lfloor \frac{2005}{3} \right\rfloor = 668, \qquad \left\lfloor \frac{2005}{4} \right\rfloor = 501, \qquad \left\lfloor \frac{2005}{12} \right\rfloor = 167.$

A multiple of $12$ is a multiple of both $3$ and $4$ , so it appears in both of the first two counts. We want such numbers removed completely, so we must take them out twice:
$668 + 501 - 2 \cdot 167 = 1169 - 334 = 835.$

The answer is $\boxed{\textbf{(C)}\ 835}$ .

Why this works

Standard inclusion-exclusion subtracts the overlap once to get "multiples of $3$ or $4$ ." Excluding the overlap altogether requires one more subtraction. Think of it as a Venn diagram: we want the two crescents but not the lens in the middle, and the lens was counted twice by the raw sums.

Alternative approach

Every block of $12$ consecutive integers contains exactly $5$ qualifying numbers (residues $3, 4, 6, 8, 9$ ). There are $167$ full blocks in $1$ to $2004$ , giving $835$ , and $2005$ (residue $1$ ) adds nothing.

The trap

Subtracting the multiples of 12 only once (ordinary inclusion-exclusion), which leaves them in and gives 1002.

Common mistakes

Subtracting the multiples of 12 only once (ordinary inclusion-exclusion), which leaves them in and gives 1002.
Not subtracting at all ( $1169$ , choice (E)) or miscomputing one of the floor quotients.

Techniques

Set up the equation/formula and compute; no special trick needed