2020 AMC 10A Problems/Problem 22

For how many positive integers $n \le 1000$ is $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$ not divisible by $3$ ? (Recall that $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ .)

$\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26$

Solution

Let $a = \lfloor \frac{998}n \rfloor$ . If the expression is not divisible by $3$ , then the three terms in the expression must be $(a, a + 1, a + 1)$ , which would imply that $n$ is divisible by $999$ but not $1000$ , or $(a, a, a + 1)$ , which would imply that $n$ is divisible by $1000$ but not $999$ . $999 = 3^3 \cdot 37$ has $4 \cdot 2 = 8$ factors, and $1000 = 2^3 \cdot 5^3$ has $4 \cdot 4 = 16$ factors. However, $n = 1$ does not work because $1$ is divisible by both $999$ and $1000$ , and since $1$ is counted twice, the answer is $16 + 8 - 2 = \boxed{\textbf{(A) }22}$ .

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2020 AMC 10A Problems/Problem 22

Solution

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