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

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==Problem==
 
For how many positive integers <math>n \le 1000</math> is<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>not divisible by <math>3</math>? (Recall that <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.)
 
For how many positive integers <math>n \le 1000</math> is<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>not divisible by <math>3</math>? (Recall that <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.)
  
 
<math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math>
 
<math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math>
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== Solution 1 (Casework) ==
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<b>Expression:</b>
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<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>
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<b>Solution:</b>
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Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>
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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.
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This means that for every integer <math>n \geq 2</math>,
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<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>,
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<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>,
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<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>,
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<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
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<math>\bullet</math> if none of <math>\left\{\frac{998}n, \frac{999}n, \frac{1000}n\right\}</math> are integral, then the three terms in the expression above must be <math>(a, a, a)</math>.
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The last statement is true because in order for the terms to be different, there must be some integer in the interval <math>\left(\frac{998}n, \frac{999}n\right)</math> or the interval <math>\left(\frac{999}n, \frac{1000}n\right)</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.
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<math>\bullet</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.
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This gives us
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<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.
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Now, we test the five cases listed above (where <math>n \geq 2</math>)
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<b>Case 1:</b> <math>n</math> divides <math>998</math> and <math>n \neq 2</math>
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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>.
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Therefore, the first case does not work (<b>0</b> cases).
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<b>Case 2:</b> <math>n</math> divides <math>998</math> and <math>n = 2</math>
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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 <b>1</b> case that works.
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<b>Case 3:</b> <math>n</math> divides <math>999</math>
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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>.
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<math>999</math> = <math>3^3 \cdot 37^1</math>.
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So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</math>.
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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 =</math> <b>7</b> cases.
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<b>Case 4:</b> <math>n</math> divides <math>1000</math>
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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>.
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<math>1000</math> = <math>5^3 \cdot 2^3</math>.
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So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</math>.
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Again, we have to subtract <math>1</math>, so this leaves <math>16 - 1 = 15</math> cases.
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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 (Case 2).
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<math>15 - 1 = 14</math>, so there are actually <b>14</b> valid cases.
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<b>Case 5:</b> <math>n</math> divides none of <math>\{998, 999, 1000\}</math>
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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 (<b>0</b> cases).
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Now that we have counted all of the cases, we add them.
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<math>0 + 1 + 7 + 14 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
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~dragonchomper, additional edits by emerald_block
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== Solution 2 (Solution 1 but simpler) ==
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Notice that you only need to count the number of factors of <math>1000</math> and <math>999</math>, excluding <math>1</math>.
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<math>1000</math> has <math>16</math> factors, and <math>999</math> has <math>8</math>.
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Adding them gives you <math>24</math>, but you need to subtract <math>2</math> since <math>1</math> does not work.
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Therefore, the answer is <math>24 - 2 = \boxed{\textbf{(A)}22}</math>.
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-happykeeper, additional edits by dragonchomper, even more edits by ericshi1685
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== Solution 3 - Solution 1 but much simpler ==
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NOTE: For this problem, whenever I say <math>\text{*factors*}</math>, I will be referring to all the factors of the number except for <math>1</math>.
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Now, quickly observe that if <math>n>2</math> divides <math>998</math>, then <math>\left\lfloor {\frac{999}{n}} \right\rfloor</math> and <math>\left\lfloor {\frac{1000}{n}} \right\rfloor</math> will also round down to <math>\frac{998}{n}</math>, giving us a sum of <math>3 \cdot \frac{998}{n}</math>, which does not work for the question. However, if <math>n>2</math> divides <math>999</math>, we see that <math>\left\lfloor {\frac{998}{n}} \right\rfloor = \frac{999}{n}-1</math> and <math>\left\lfloor {\frac{1000}{n}} \right\rfloor=\left\lfloor {\frac{999}{n}} \right\rfloor</math>. This gives us a sum of <math>3 \cdot \left\lfloor {\frac{999}{n}} \right\rfloor - 1</math>, which is clearly not divisible by <math>3</math>. Using the same logic, we can deduce that <math>(n>2)|1000</math> works, too (for our problem). Thus, we need the factors of <math>999</math> and <math>1000</math> and we don't have to eliminate any because the <math>\text{gcf} (999,1000)=1</math>. But we have to be careful! See that when <math>n|998,999,1000</math>, then our problem doesn't get fulfilled. The only <math>n</math> that satisfies that is <math>n=1</math>. So, we have:
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<math>999=3^3\cdot 37 \implies (3+1)(1+1)-1 \text{*factors*} \implies 7</math>;
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<math>1000=2^3\cdot 5^3 \implies (3+1)(3+1)-1 \text{*factors*} \implies 15</math>.
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Adding them up gives a total of <math>7+15=\boxed{\textbf{(A)}22}</math> workable <math>n</math>'s.
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== Solution 4 ==
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Writing out <math>n = 1, 2, 3, 4 ... 11</math>, we see that the answer cannot be more than the number of divisors of <math>998, 999, 1000</math> since all <math>n</math> satisfying the problem requirements are among the divisors of <math>998, 999, 1000</math>. There are <math>28</math> total divisors, and we subtract <math>3</math> from the start because we count <math>1</math>, which never works, thrice.
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From the divisors of <math>998</math>, note that <math>499</math> and <math>998</math> don't work. 2 to subtract.
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Also note that we count <math>2</math> twice, in <math>998</math> and <math>1000</math>, so we have to subtract another from the running total of <math>25</math>.
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Already, we are at <math>22</math> divisors so we conclude that the answer is <math>\boxed{\textbf{(A)}22}</math>.
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== Solution 5 ==
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First, we notice the following lemma:
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<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>
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<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>
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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>
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From the lemma and the given equation, we have four possible cases:
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<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>
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<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>
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<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>
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<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>
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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 n's for which cases (2) and (3) stand.
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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.
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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.
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So in total, we have total of <math>15+7=\boxed{\textbf{(A)}22}</math> possible <math>n</math>'s.
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~mathboywannabe
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==Video Solution 1==
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https://www.youtube.com/watch?v=_Ej9nnHS07s
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~Snore
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==Video Solution 2==
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Education, The Study of Everything
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https://youtu.be/LWAYKQQX6KI
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 +
==Video Solution 3==
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https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx
  
 
==See Also==
 
==See Also==

Latest revision as of 12:43, 9 January 2021

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

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 $\left\{\frac{998}n, \frac{999}n, \frac{1000}n\right\}$ 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 $\left(\frac{998}n, \frac{999}n\right)$ or the interval $\left(\frac{999}n, \frac{1000}n\right)$. 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.


$\bullet$ 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 (0 cases).


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 (Case 2). $15 - 1 = 14$, 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 (0 cases).


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

Solution 2 (Solution 1 but simpler)

Notice that you only need to count the number of factors of $1000$ and $999$, excluding $1$. $1000$ has $16$ factors, and $999$ has $8$. Adding them gives you $24$, but you need to subtract $2$ since $1$ does not work.

Therefore, the answer is $24 - 2 = \boxed{\textbf{(A)}22}$.

-happykeeper, additional edits by dragonchomper, even more edits by ericshi1685

Solution 3 - Solution 1 but much simpler

NOTE: For this problem, whenever I say $\text{*factors*}$, I will be referring to all the factors of the number except for $1$.

Now, quickly observe that if $n>2$ divides $998$, then $\left\lfloor {\frac{999}{n}} \right\rfloor$ and $\left\lfloor {\frac{1000}{n}} \right\rfloor$ will also round down to $\frac{998}{n}$, giving us a sum of $3 \cdot \frac{998}{n}$, which does not work for the question. However, if $n>2$ divides $999$, we see that $\left\lfloor {\frac{998}{n}} \right\rfloor = \frac{999}{n}-1$ and $\left\lfloor {\frac{1000}{n}} \right\rfloor=\left\lfloor {\frac{999}{n}} \right\rfloor$. This gives us a sum of $3 \cdot \left\lfloor {\frac{999}{n}} \right\rfloor - 1$, which is clearly not divisible by $3$. Using the same logic, we can deduce that $(n>2)|1000$ works, too (for our problem). Thus, we need the factors of $999$ and $1000$ and we don't have to eliminate any because the $\text{gcf} (999,1000)=1$. But we have to be careful! See that when $n|998,999,1000$, then our problem doesn't get fulfilled. The only $n$ that satisfies that is $n=1$. So, we have: $999=3^3\cdot 37 \implies (3+1)(1+1)-1 \text{*factors*} \implies 7$; $1000=2^3\cdot 5^3 \implies (3+1)(3+1)-1 \text{*factors*} \implies 15$. Adding them up gives a total of $7+15=\boxed{\textbf{(A)}22}$ workable $n$'s.

Solution 4

Writing out $n = 1, 2, 3, 4 ... 11$, we see that the answer cannot be more than the number of divisors of $998, 999, 1000$ since all $n$ satisfying the problem requirements are among the divisors of $998, 999, 1000$. There are $28$ total divisors, and we subtract $3$ from the start because we count $1$, which never works, thrice.

From the divisors of $998$, note that $499$ and $998$ don't work. 2 to subtract. Also note that we count $2$ twice, in $998$ and $1000$, so we have to subtract another from the running total of $25$.

Already, we are at $22$ divisors so we conclude that the answer is $\boxed{\textbf{(A)}22}$.

Solution 5

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

Video Solution 1

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

~Snore

Video Solution 2

Education, The Study of Everything

https://youtu.be/LWAYKQQX6KI

Video Solution 3

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

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

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