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

Since $\frac{1000}n - \frac{998}n = \frac{2}n$ , for any integer $n \geq 2$ , the difference between the largest and smallest terms before the $\lfloor x \rfloor$ function is applied is less than or equal to $1$ , and thus the terms must have a range of $1$ or less after the function is applied.

This means that for every integer $n \geq 2$ ,

$\bullet$ if $\frac{998}n$ is an integer and $n \neq 2$ , then the three terms in the expression above must be $(a, a, a)$ ,

$\bullet$ if $\frac{998}n$ is an integer because $n = 2$ , then $\frac{1000}n$ will be an integer and will be $1$ greater than $\frac{998}n$ ; thus the three terms in the expression must be $(a, a, a + 1)$ ,

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

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

$\bullet$ if none of $\{\frac{998}n, \frac{999}n, \frac{1000}n\}$ are integral, then the three terms in the expression above must be $(a, a, a)$ .

The last statement is true because in order for the terms to be different, there must be some integer in the interval $(\frac{998}n, \frac{999}n)$ or the interval $(\frac{999}n, \frac{1000}n)$ . However, this means that multiplying the integer by $n$ should produce a new integer between $998$ and $999$ or $999$ and $1000$ , exclusive, but because no such integers exist, the terms cannot be different, and thus, must be equal.

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

Now, we test the five cases listed above (where $n \geq 2$ )

Case 1: $n$ divides $998$ and $n \neq 2$

As mentioned above, the three terms in the expression are $(a, a, a)$ , so the sum is $3a$ , which is divisible by $3$ . Therefore, the first case does not work.

Case 2: $n$ divides $998$ and $n = 2$

As mentioned above, in this case the terms must be $(a, a, a + 1)$ , which means the sum is $3a + 1$ , so the expression is not divisible by $3$ . Therefore, this is $1$ case that works.

Case 3: $n$ divides $999$

Because $n$ divides $999$ , the number of possibilities for $n$ is the same as the number of factors of $999$ .

$999$ = $3^3 \cdot 37^1$ . So, the total number of factors of $999$ is $4 \cdot 2 = 8$ .

However, we have to subtract $1$ , because the case $n = 1$ does not work, as mentioned previously. This leaves $8 - 1 = 7$ cases.

Case 4: $n$ divides $1000$

Because $n$ divides $1000$ , the number of possibilities for $n$ is the same as the number of factors of $1000$ .

$1000$ = $5^3 \cdot 2^3$ . So, the total number of factors of $1000$ is $4 \cdot 4 = 16$ .

Again, we have to subtract $1$ , so this leaves $16 - 1 = 15$ cases. We have also overcounted the factor $2$ , as it has been counted as a factor of $1000$ and as a separate case. $15 - 1 = 14$ , so there are actually $14$ valid cases.

Case 5: $n$ divides none of $\{998, 999, 1000\}$

Similar to Case 1, the value of the terms of the expression are $(a, a, a)$ . The sum is $3a$ , which is divisible by 3, so this case does not work.

Now that we have counted all of the cases, we add them.

$0 + 1 + 7 + 14 + 0 = 22$ , so the answer is $\boxed{\textbf{(A)}22}$ .

~dragonchomper, additional edits by emerald_block

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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