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

(→‎Solution (Casework))
(fixed mistakes, elaborated more)
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Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>
 
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>
  
−
Notice that for every integer <math>n \neq 1</math>,
+
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.
  
−
<math>\bullet</math> if <math>\frac{998}n</math> is an integer, then the three terms in the expression above must be <math>(a, a, a)</math>,
+
This means that for every integer <math>n \geq 2</math>,
 +
 
 +
<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>,
  
 
<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>,
 
<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>,
  
−
<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
+
<math>\bullet</math> if <math>\frac{998}n</math> is an integer and <math>n \neq 2</math> (since if <math>n = 2</math>, <math>\frac{1000}n</math> will be an integer, and it will be <math>1</math> greater than <math>\frac{998}n</math>), then the three terms in the expression above must be <math>(a, a, a)</math>, and
  
−
<math>\bullet</math> if none of {<math>\frac{998}n</math>, <math>\frac{999}n</math>, <math>\frac{1000}n</math>} are integral, then the three terms in the expression above must be <math>(a, a, a)</math>.
+
<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>.
  
−
This is due to the fact that <math>998</math>, <math>999</math>, and <math>1000</math> do not collectively share any common factors (other than 1).
+
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 original integer cannot exist, and thus, the terms must be equal.
  
  
−
Note that <math>n = 1</math> doesn't work; to prove this, we just have to substitute <math>1</math> for <math>n</math> in the expression.
+
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.
 
This gives us
 
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 = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3</math> which is divisible by 3.
+
<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.
−
Therefore, the case <math>n = 1</math> does not work.
 
  
  
Line 39: Line 40:
 
<b>Case 1:</b> <math>n</math> divides <math>998</math>
 
<b>Case 1:</b> <math>n</math> divides <math>998</math>
  
−
The three terms in the expression must be <math>(a, a, a)</math>, as mentioned above, so the sum is <math>3a</math>, which is divisible by <math>3</math>.
+
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>.
 
Therefore, the first case does not work.
 
Therefore, the first case does not work.
  
Line 50: Line 51:
 
So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</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> doesn't work, as mentioned previously.
+
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.
−
<math>8 - 1 = 7</math>
 
−
 
 
−
We now do the same for the third case.
 
  
  
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So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</math>.
 
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>, for the reason stated in Case 2.
+
Again, we have to subtract <math>1</math>, so this leaves <math>16 - 1 = 15</math> cases.
−
<math>16 - 1 = 15</math>
 
−
 
 
−
 
 
−
<b>Case 4:</b> <math>n</math> divides none of {<math>998, 999, 1000</math>}
 
  
−
If <math>n</math> does not divide {<math>998, 999, 1000</math>}, then the value of the terms of the above expression are <math>(a, a, a)</math>.
 
−
See if you can figure out why, although the explanation is similar to that of Case 1.
 
  
−
Therefore, the value of the expression is <math>3a</math>, which is divisible by 3, so this case does not work.
+
<b>Case 4:</b> <math>n</math> divides none of <math>\{998, 999, 1000\}</math>
  
 +
As in 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.
  
  
Line 80: Line 73:
 
<math>0 + 7 + 15 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
 
<math>0 + 7 + 15 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
  
−
~dragonchomper
+
~dragonchomper, additional edits by [[User:emerald_block|emerald_block]]
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:35, 2 February 2020

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 $\frac{1000}n$ is an integer, then the three terms in the expression above 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 $\frac{998}n$ is an integer and $n \neq 2$ (since if $n = 2$, $\frac{1000}n$ will be an integer, and it will be $1$ greater than $\frac{998}n$), then the three terms in the expression above must be $(a, a, a)$, 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 original integer cannot exist, and thus, the terms 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 four cases listed above.


Case 1: $n$ divides $998$

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 $999$

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 3: $n$ 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.


Case 4: $n$ divides none of $\{998, 999, 1000\}$

As in 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.

$0 + 7 + 15 + 0 = 22$, so the answer is $\boxed{\textbf{(A)}22}$.

~dragonchomper, additional edits by emerald_block

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

See Also

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
All AMC 10 Problems and Solutions

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