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 1 (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 other than 1.

Note that $n = 1$ doesn't work; to prove this, we just have to substitute $1$ for $n$ in the expression, to get

$\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor$

This gives us $\frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3$ which is divisible by 3. Therefore, the case $n = 1$ does not work.

Now, we test the three cases mentioned above.

Case 1: $n$ 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: $n$ 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 $1$ , because the case $n = 1$ doesn't work, as mentioned previously.

$8 - 1 = 7$

We now do the same for the third case.

Case 3: $n$ divides $1000$

$1000$ = $5^3 \cdot 2^3$

So, the total number of factors of $1000$ is $4 \cdot 4 = 16$ .

Again, we have to subtract $1$ , 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
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2020 AMC 10A Problems/Problem 22

Contents

Problem

Solution 1 (Casework)

Video Solution

See Also