Difference between revisions of "2020 AMC 10A Problems/Problem 22"

Latest revision as of 21:13, 29 September 2023

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

Clearly, $n=1$ fails. Except for the special case of $n=1$ , $\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor$ equals either $0$ or $1$ . If it equals $0$ , this implies that $\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor$ , so their sum is clearly a multiple of $3$ , so this will always fail. If it equals $1$ , the sum of the three floor terms is $3 \left\lfloor \frac{999}{n} \right\rfloor \pm 1$ , so it is never a multiple of $3$ . Thus, we are looking for all $n \neq 1$ such that $\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1.$ This implies that either $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor,$ or $\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor.$

Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer $a$ such that $\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999.*$ Analogously, the second equation implies that $n | 1000.$ So our only $n$ that satisfy this condition are $n \neq 1$ that divide $999$ or $1000$ . Using the method to find the number of divisors of a number, we see that $999$ has $8$ divisors and $1000$ has $16$ divisors. Their only common factor is $1$ , so there are $8+16-1 = 23$ positive integers that divide either $999$ or $1000$ . Since the integer $1$ is a special case and does not count, we must subtract this from our $23$ , so our final answer is $23-1 = \boxed{\textbf{(A) } 22}.$

*While this observation may seem strange, it is actually "trivial by intuition" to go straight from  to . In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.

~ihatemath123

Solution 2

Counting down $n$ from $1000$ , $999$ , $998$ ... we notice that $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$ is only not divisible by $3$ when n is a divisor of only $1000$ or $999 (1000, 999, 500, 333$ ...). Notice how the factors of $998: 1, 2, 499$ , and $998$ , do not work.

The prime factorization of $999$ is $3^3\cdot37$ , so $4\cdot2=8$ factors in total.

The prime factorization of $1000$ is $2^3\cdot5^3$ , so $4\cdot4=16$ factors in total.

However, $1$ obviously does not work, so we have to subtract $2$ ( $1$ is counted twice) from the total. $8 + 16 - 2$ = $\boxed{\textbf{(A)}22}$ .

~BLOATED_BAGEL

~Typo fixed by evanhliu2009

Solution 3

First, we notice the following lemma:

$\textbf{Lemma}$ : For $N, n \in \mathbb{N}$ , $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1$ if $n \mid N$ ; and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor$ if $n \nmid N.$

$\textbf{Proof}$ : Let $A = kn + r$ , with $0 \leq r < n$ . If $n \mid N$ , then $r = 0$ . Hence $\left\lfloor \frac{N}{n} \right\rfloor = k$ , $\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1$ , and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1.$

If $n \nmid N$ , then $1 \leq r < n$ . Hence $\left\lfloor \frac{N}{n} \right\rfloor = k$ , $\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k$ , and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor.$

From the lemma and the given equation, we have four possible cases: $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1 \qquad (1)$ $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (2)$ $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (3)$ $\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (4)$

Note that cases (2) and (3) are the cases in which the term, $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor,$ is not divisible by $3$ . So we only need to count the number of $n$ 's for which cases (2) and (3) stand.

Case (2): By the lemma, we have $n \mid 1000$ and $n \nmid 999.$ Hence $n$ can be any factor of $1000$ except for $n = 1$ . Since $1000 = 2^3 * 5^3,$ there are $(3+1)(3+1) - 1 = 15$ possible values of $n$ for this case.

Case (3): By the lemma, we have $n \mid 999$ and $n \nmid 998.$ Hence $n$ can be any factor of $999$ except for $n = 1$ . Since $999 = 3^3 * 37^1,$ there are $(3+1)(1+1) - 1 = 7$ possible values of $n$ for this case.

So in total, we have total of $15+7=\boxed{\textbf{(A)}22}$ possible $n$ 's.

~mathboywannabe

Solution 4 (Casework)

Note that $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor$ is a multiple of $3$ if $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor$ lies between two consecutive multiples of $n$ .

$\textbf{Explanation:}$

Let's assume that the above expression does indeed lie betweent two consecutive multiples of $n$ . This implies that $n$ does not divide either $998, 999$ or $1000$ , meaning that when divided by $n$ , none of the quotients are whole. In turn, this also means that they will all have the same whole number part.

If $\frac{998}{n}, \frac{999}{n},$ and $\frac{1000}{n}$ were to have a different whole number part, then $n$ would have to lie within $998$ and $1000$ . If this is confusing, think of why $\frac{6}{5}, \frac{7}{5}$ and $\frac{8}{5}$ have the same whole number part ( $1$ in this case). Here, $n=5$ . What happens to the whole number part when $n=7$ ?

Since $\frac{998}{n}, \frac{999}{n},$ and $\frac{1000}{n}$ have identical whole number parts, their floors are identical, since the floor of a number is equal to its whole number part, discarding its fractional component.

Let $k$ be the number we get when we floor $\frac{998}{n}, \frac{999}{n},$ or $\frac{1000}{n}$ if the three of those numbers lie between two consecutive multiples of $n$ . Adding them up, we get $3k$ (due to their floors being the same), which is a mulutiple of $3$ . So no matter what, we cannot have $\frac{998}{n}, \frac{999}{n},$ and $\frac{1000}{n}$ lie between two consecutive multiples of $n$ .

What does this mean? It means that there must be some multiple of $n$ within this expression (with some restrictions, as we'll see), in order to prevent the violation of the main restriction. We can proceed with casework now.

$\textbf{Case 1}$

Let's assume that the multiple of $n$ is located at $\frac{998}{n}$ . Luckily, the only prime factors of $998$ are $2$ and $499$ .

We can observe that $499$ doesn't work, since $\frac{998}{499}$ and $\frac{1000}{499}$ will both round down to $2$ when divided by it. However, $2$ does work, since it divides both $998$ and $1000$ . The floor of $\frac{999}{2}$ is $499$ , , meaning that when we evaluate the given expression for $n=2$ , we will get $2\cdot{499}+500$ which isn't a multiple of $3$ .

This case works because we have a multiple of $n$ at the end of the expression, as well as the beginning, so the whole number parts of $998, 999$ and $1000$ when divided by $n$ are not all the same, due to $n$ dividing two of these numbers.

The restriction of $n$ dividing more than one number within the expression is only valid when we're testing the first number in the given expression (otherwise, the other floors wouldn't round down to the same value as the first).

Case $1$ is complete, and we've found that only $2$ works.

$\textbf{Case 2}$

Now, let's assume that the multiple of $n$ is located at $\frac{999}{n}$ . In this case, if $n$ divides $999$ , it doesn't divide $998$ (since two multiples of a number greater than $1$ are never consecutive), nor does it divide $1000$ , for the same reason.

The prime factorization of $999$ is $3^3\cdot37$ , and thus has $(3+1)(1+1)=8$ divisors.

When testing a few values of $n$ initially, we observed that $n=1$ causes the expression to be divisible by $3$ . Subtracting $1$ , we see that there are $7$ values of $n$ that work for this case.

$\textbf{Case 3}$

Finally, let's assume that the multiple of $n$ is located at $\frac{1000}{n}.$

Our goal is to have the other multiple of $n$ below whatever $\frac{998}{n}$ is, so we won't have identical floors throughout the expression.

Once again, any factor of $1000$ , except for $2$ is relatively prime to both $998$ and $999$ , so when we floor those two numbers, we get an integer that isn't $500$ .

$1000$ factors as $2^3\cdot{5^3}$ , meaning that it has $(3+1)(3+1)=16$ divisors. $1$ and $2$ either don't work or have already been counted, so there are $14$ valid values of $n$ for this case.

$\textbf{Ending}$

Adding these three cases, we get $1+7+14= \boxed{\textbf{A) }22}$ values of $n$ .

-Benedict T (countmath1)

Video Solutions

Video Solution 1 (Simple)

Education, The Study of Everything

https://youtu.be/LWAYKQQX6KI

Video Solution 2

https://www.youtube.com/watch?v=_Ej9nnHS07s

~Snore

Video Solution 3

https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx

Video Solution 4 (Richard Rusczyk)

https://artofproblemsolving.com/videos/amc/2020amc10a/517

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math>
+== Solution 1 ==
+Clearly, <math>n=1</math> fails. Except for the special case of <math>n=1</math>,
+<cmath>\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor</cmath>
+equals either <math>0</math> or <math>1</math>. If it equals <math>0</math>, this implies that <math>\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor</math>, so their sum is clearly a multiple of <math>3</math>, so this will always fail. If it equals <math>1</math>, the sum of the three floor terms is <math>3 \left\lfloor \frac{999}{n} \right\rfloor \pm 1</math>, so it is never a multiple of <math>3</math>. Thus, we are looking for all <math>n \neq 1</math> such that
+<cmath>\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1.</cmath>
+This implies that either
+<cmath>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor,</cmath>
+or
+<cmath>\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor.</cmath>
-== Solution (Casework) ==
+Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer <math>a</math> such that
+<cmath>\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999.*</cmath>
+Analogously, the second equation implies that
+<cmath>n | 1000.</cmath>
+So our only <math>n</math> that satisfy this condition are <math>n \neq 1</math> that divide <math>999</math> or <math>1000</math>. Using the method to find the number of divisors of a number, we see that <math>999</math> has <math>8</math> divisors and <math>1000</math> has <math>16</math> divisors. Their only common factor is <math>1</math>, so there are <math>8+16-1 = 23</math> positive integers that divide either <math>999</math> or <math>1000</math>. Since the integer <math>1</math> is a special case and does not count, we must subtract this from our <math>23</math>, so our final answer is <math>23-1 = \boxed{\textbf{(A) } 22}.</math>
-<b>Expression:</b>
+ *While this observation may seem strange, it is actually "trivial by intuition" to go straight from <math>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor</math> to <math>n | 999</math>. In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.
-<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>
-<b>Solution:</b>
+~ihatemath123
-Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>
+== Solution 2 ==
-Since <math>\frac{1000}n - \frac{998}n = \frac{2}n</math>, for any integer <math>n \geq 2</math>, the difference between the largest and smallest terms before the <math>\lfloor x \rfloor</math> function is applied is less than or equal to <math>1</math>, and thus the terms must have a range of <math>1</math> or less after the function is applied.
+Counting down <math>n</math> from <math>1000</math>, <math>999</math>, <math>998</math>... we notice that <math>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</math> is only not divisible by <math>3</math> when n is a divisor of ''only'' <math>1000</math> or <math>999 (1000, 999, 500, 333</math>...). Notice how the factors of <math>998: 1, 2, 499</math>, and <math>998</math>, do not work.
-This means that for every integer <math>n \geq 2</math>,
+The prime factorization of <math>999</math> is <math>3^3\cdot37</math>, so <math>4\cdot2=8</math> factors in total.
-<math>\bullet</math> if <math>\frac{998}n</math> is an integer and <math>n \neq 2</math>, then the three terms in the expression above must be <math>(a, a, a)</math>,
+The prime factorization of <math>1000</math> is <math>2^3\cdot5^3</math>, so <math>4\cdot4=16</math> factors in total.
-<math>\bullet</math> if <math>\frac{998}n</math> is an integer because <math>n = 2</math>, then <math>\frac{1000}n</math> will be an integer and will be <math>1</math> greater than <math>\frac{998}n</math>; thus the three terms in the expression must be <math>(a, a, a + 1)</math>,
+However, <math>1</math> obviously does not work, so we have to subtract <math>2</math> (<math>1</math> is counted twice) from the total. <math>8 + 16 - 2</math> = <math>\boxed{\textbf{(A)}22}</math>.
-<math>\bullet</math> if <math>\frac{999}n</math> is an integer, then the three terms in the expression above must be <math>(a, a + 1, a + 1)</math>,
+~BLOATED_BAGEL
-<math>\bullet</math> if <math>\frac{1000}n</math> is an integer, then the three terms in the expression above must be <math>(a, a, a + 1)</math>, and
+~Typo fixed by evanhliu2009
-<math>\bullet</math> if none of <math>\{\frac{998}n, \frac{999}n, \frac{1000}n\}</math> are integral, then the three terms in the expression above must be <math>(a, a, a)</math>.
+== Solution 3 ==
+First, we notice the following lemma:
-The last statement is true because in order for the terms to be different, there must be some integer in the interval <math>(\frac{998}n, \frac{999}n)</math> or the interval <math>(\frac{999}n, \frac{1000}n)</math>. However, this means that multiplying the integer by <math>n</math> should produce a new integer between <math>998</math> and <math>999</math> or <math>999</math> and <math>1000</math>, exclusive, but because no such integers exist, the terms cannot be different, and thus, must be equal.
+<math>\textbf{Lemma}</math>: For <math>N, n \in \mathbb{N}</math>, <math> \left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1 </math> if <math>n \mid N</math>; and <math>\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor</math>  if <math>n \nmid N.</math>
+<math>\textbf{Proof}</math>: Let <math>A = kn + r</math>, with <math>0 \leq r < n</math>. If <math>n \mid N</math>, then <math>r = 0</math>. Hence <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math>, <math>\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1</math>, and <math>\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1.</math>
-Note that <math>n = 1</math> does not work; to prove this, we just have to substitute <math>1</math> for <math>n</math> in the expression.
+If <math>n \nmid N</math>, then <math>1 \leq r < n</math>. Hence <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math>, <math>\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k</math>, and <math>\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor.</math>
-This gives us
-<math>\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</math> which is divisible by 3.
+From the lemma and the given equation, we have four possible cases:
+<cmath>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1 \qquad (1)</cmath>
+<cmath>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (2)</cmath>
+<cmath>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (3)</cmath>
+<cmath>\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (4)</cmath>
-Now, we test the five cases listed above (where <math>n \geq 2</math>)
+Note that cases (2) and (3) are the cases in which the term, <math>\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor,</math> is not divisible by <math>3</math>. So we only need to count the number of <math>n</math>'s for which cases (2) and (3) stand.
+Case (2): By the lemma, we have <math>n \mid 1000</math> and <math>n \nmid 999.</math> Hence <math>n</math> can be any factor of <math>1000</math> except for <math>n = 1</math>. Since <math>1000 = 2^3 * 5^3,</math> there are <math>(3+1)(3+1) - 1 = 15</math> possible values of <math>n</math> for this case.
-<b>Case 1:</b> <math>n</math> divides <math>998</math> and <math>n \neq 2</math>
+Case (3): By the lemma, we have <math>n \mid 999</math> and <math>n \nmid 998.</math> Hence <math>n</math> can be any factor of <math>999</math> except for <math>n = 1</math>. Since <math>999 = 3^3 * 37^1,</math> there are <math>(3+1)(1+1) - 1 = 7</math> possible values of <math>n</math> for this case.
-As mentioned above, the three terms in the expression are <math>(a, a, a)</math>, so the sum is <math>3a</math>, which is divisible by <math>3</math>.
+So in total, we have total of <math>15+7=\boxed{\textbf{(A)}22}</math> possible <math>n</math>'s.
-Therefore, the first case does not work.
-<b>Case 2:</b> <math>n</math> divides <math>998</math> and <math>n = 2</math>
+~mathboywannabe
-As mentioned above, in this case the terms must be <math>(a, a, a + 1)</math>, which means the sum is <math>3a + 1</math>, so the expression is not divisible by <math>3</math>. Therefore, this is <math>1</math> case that works.
+== Solution 4 (Casework)==
+Note that <math>\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor</math> is a multiple of <math>3</math> if <math>\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor</math> lies between two consecutive multiples of <math>n</math>.
-<b>Case 3:</b> <math>n</math> divides <math>999</math>
-Because <math>n</math> divides <math>999</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>999</math>.
+<math>\textbf{Explanation:}</math>
-<math>999</math> = <math>3^3 \cdot 37^1</math>.
-So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</math>.
-However, we have to subtract <math>1</math>, because the case <math>n = 1</math> does not work, as mentioned previously. This leaves <math>8 - 1 = 7</math> cases.
+Let's assume that the above expression does indeed lie betweent two consecutive multiples of <math>n</math>. This implies that <math>n</math> does not divide either <math>998, 999</math> or <math>1000</math>, meaning that when divided by <math>n</math>, none of the quotients are whole. In turn, this also means that they will all have the same whole number part.
-<b>Case 4:</b> <math>n</math> divides <math>1000</math>
+If <math>\frac{998}{n}, \frac{999}{n},</math> and <math>\frac{1000}{n}</math> were to have a different whole number part, then <math>n</math> would have to lie within <math>998</math> and <math>1000</math>. If this is confusing, think of why <math>\frac{6}{5}, \frac{7}{5}</math> and <math>\frac{8}{5}</math> have the same whole number part (<math>1</math> in this case). Here, <math>n=5</math>. What happens to the whole number part when <math>n=7</math>?
-Because <math>n</math> divides <math>1000</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>1000</math>.
-<math>1000</math> = <math>5^3 \cdot 2^3</math>.
+Since <math>\frac{998}{n}, \frac{999}{n},</math> and <math>\frac{1000}{n}</math> have identical whole number parts, their floors are identical, since the floor of a number is equal to its whole number part, discarding its fractional component.
-So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</math>.
-Again, we have to subtract <math>1</math>, so this leaves <math>16 - 1 = 15</math> cases.
-We have also overcounted the factor <math>2</math>, as it has been counted as a factor of <math>1000</math> and as a separate case.
-<math>15 - 1 = 14</math>, so there are actually <math>14</math> valid cases.
+Let <math>k</math> be the number we get when we floor <math>\frac{998}{n}, \frac{999}{n},</math> or <math>\frac{1000}{n}</math> if the three of those numbers lie between two consecutive multiples of <math>n</math>. Adding them up, we get <math>3k</math> (due to their floors being the same), which is a mulutiple of <math>3</math>. So no matter what, we cannot have <math>\frac{998}{n}, \frac{999}{n},</math> and <math>\frac{1000}{n}</math> lie between two consecutive multiples of <math>n</math>.
-<b>Case 5:</b> <math>n</math> divides none of <math>\{998, 999, 1000\}</math>
-Similar to Case 1, the value of the terms of the expression are <math>(a, a, a)</math>. The sum is <math>3a</math>, which is divisible by 3, so this case does not work.
+What does this mean? It means that there must be some multiple of <math>n</math> within this expression (with some restrictions, as we'll see), in order to prevent the violation of the main restriction. We can proceed with casework now.
-Now that we have counted all of the cases, we add them.
+<math>\textbf{Case 1}</math>
-<math>0 + 1 + 7 + 14 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
-~dragonchomper, additional edits by emerald_block
+Let's assume that the multiple of <math>n</math> is located at <math>\frac{998}{n}</math>. Luckily, the only prime factors of <math>998</math> are <math>2</math> and <math>499</math>.
-==Video Solution==
-https://youtu.be/Ozp3k2464u4
-~IceMatrix
+We can observe that <math>499</math> doesn't work, since <math>\frac{998}{499}</math> and <math>\frac{1000}{499}</math> will both round down to <math>2</math> when divided by it. However, <math>2</math> does work, since it divides both <math>998</math> and <math>1000</math>. The floor of <math>\frac{999}{2}</math> is <math>499</math>, , meaning that when we evaluate the given expression for <math>n=2</math>, we will get <math>2\cdot{499}+500</math> which isn't a multiple of <math>3</math>.
+This case works because we have a multiple of <math>n</math> at the end of the expression, as well as the beginning, so the whole number parts of <math>998, 999</math> and <math>1000</math> when divided by <math>n</math> are not all the same, due to <math>n</math> dividing two of these numbers.
+The restriction of <math>n</math> dividing more than one number within the expression is only valid when we're testing the first number in the given expression (otherwise, the other floors wouldn't round down to the same value as the first).
+Case <math>1</math> is complete, and we've found that only <math>2</math> works.
+<math>\textbf{Case 2}</math>
+Now, let's assume that the multiple of <math>n</math> is located at <math>\frac{999}{n}</math>. In this case, if <math>n</math> divides <math>999</math>, it doesn't divide <math>998</math> (since two multiples of a number greater than <math>1</math> are never consecutive), nor does it divide <math>1000</math>, for the same reason.
+The prime factorization of <math>999</math> is <math>3^3\cdot37</math>, and thus has <math>(3+1)(1+1)=8</math> divisors.
+When testing a few values of <math>n</math> initially, we observed that <math>n=1</math> causes the expression to be divisible by <math>3</math>. Subtracting <math>1</math>, we see that there are <math>7</math> values of <math>n</math> that work for this case.
+<math>\textbf{Case 3}</math>
+Finally, let's assume that the multiple of <math>n</math> is located at <math>\frac{1000}{n}.</math>
+Our goal is to have the other multiple of <math>n</math> below whatever <math>\frac{998}{n}</math> is, so we won't have identical floors throughout the expression.
+Once again, any factor of <math>1000</math>, except for <math>2</math> is relatively prime to both <math>998</math> and <math>999</math>, so when we floor those two numbers, we get an integer that isn't <math>500</math>.
+<math>1000</math> factors as <math>2^3\cdot{5^3}</math>, meaning that it has <math>(3+1)(3+1)=16</math> divisors. <math>1</math> and <math>2</math> either don't work or have already been counted, so there are <math>14</math> valid values of <math>n</math> for this case.
+<math>\textbf{Ending}</math>
+Adding these three cases, we get <math>1+7+14= \boxed{\textbf{A) }22} </math> values of <math>n</math>.
+-Benedict T (countmath1)
+==Video Solutions==
+===Video Solution 1 (Simple)===
+Education, The Study of Everything
+https://youtu.be/LWAYKQQX6KI
+===Video Solution 2===
+https://www.youtube.com/watch?v=_Ej9nnHS07s
+~Snore
+===Video Solution 3===
+https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx
+===Video Solution 4 (Richard Rusczyk)===
+https://artofproblemsolving.com/videos/amc/2020amc10a/517
 ==See Also==

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