2020 AMC 10A Problems/Problem 22

Problem

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 (Casework)

Expression: $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$

Solution:

Let $a = \left\lfloor \frac{998}n \right\rfloor$

Notice that for every integer $n \neq 1$ ,

$\bullet$ if $\frac{998}n$ is an integer, then the three terms in the expression above must be $(a, a, a)$ ,

$\bullet$ if $\frac{999}n$ is an integer, then the three terms in the expression above must be $(a, a + 1, a + 1)$ , and

$\bullet$ if $\frac{1000}n$ is an integer, then the three terms in the expression above must be $(a, a, a + 1)$ .

This is due to the fact that $998$ , $999$ , and $1000$ share no common factors collectively (other than 1).

Note that $n = 1$ doesn't work; to prove this, we just have to substitute $1$ for $n$ in the expression. This gives us $\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor$ = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3 $which is divisible by 3. Therefore, the case$ n = 1$does not work.

Now, we test the three cases mentioned above.

Case 1:$ (Error compiling LaTeX. Unknown error_msg)n $divides$ 998 $The first case does not work, as the three terms in the expression must be$ (a, a, a) $, as mentioned above, so the sum becomes$ 3a $, which is divisible by$ 3$.

Case 2:$ (Error compiling LaTeX. Unknown error_msg)n $divides$ 999 $Because$ n $divides$ 999 $, the number of possibilities for$ n $is the same as the number of factors of$ 999 $, excluding$ 1 $.$ 999 $=$ 3^3 \cdot 37^1 $So, the total number of factors of$ 999 $is$ 4 \cdot 2 = 8$.

However, we have to subtract$ (Error compiling LaTeX. Unknown error_msg)1 $, because the case$ n = 1 $doesn't work, as mentioned previously.$ 8 - 1 = 7$We now do the same for the third and last case.

Case 3:$ (Error compiling LaTeX. Unknown error_msg)n $divides$ 1000 $Because$ n $divides$ 1000 $, the number of possibilities for$ n $is the same as the number of factors of$ 1000 $, excluding$ 1 $.$ 1000 $=$ 5^3 \cdot 2^3 $So, the total number of factors of$ 1000 $is$ 4 \cdot 4 = 16$.

Again, we have to subtract$ (Error compiling LaTeX. Unknown error_msg)1 $, for the reason stated in Case 2.$ 16 - 1 = 15 $Now that we have counted all of the cases, we add them.$ 7 + 15 = 22 $, so the answer is$ \boxed{\textbf{(A)}22}$.

~dragonchomper

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2020 AMC 10A Problems/Problem 22

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Solution (Casework)

Video Solution

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