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

(fixed mistakes, elaborated more)
(→‎Solution (Casework))
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This means that for every integer <math>n \geq 2</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{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>,
 +
 
 +
<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>,
  
 
<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{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 <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 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>.
 
<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>.
  
−
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.
+
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.
  
  
Line 35: Line 37:
  
  
−
Now, we test the four cases listed above.
+
Now, we test the five cases listed above (where <math>n \geq 2</math>)
  
  
−
<b>Case 1:</b> <math>n</math> divides <math>998</math>
+
<b>Case 1:</b> <math>n</math> divides <math>998</math> and <math>n \neq 2</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>.
 
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.
 +
 +
<b>Case 2:</b> <math>n</math> divides <math>998</math> and <math>n = 2</math>
 +
 +
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.
  
  
−
<b>Case 2:</b> <math>n</math> divides <math>999</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>.
 
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>.
Line 54: Line 60:
  
  
−
<b>Case 3:</b> <math>n</math> divides <math>1000</math>
+
<b>Case 4:</b> <math>n</math> divides <math>1000</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>.
 
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>.
Line 62: Line 68:
  
 
Again, we have to subtract <math>1</math>, so this leaves <math>16 - 1 = 15</math> cases.
 
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.
  
  
−
<b>Case 4:</b> <math>n</math> divides none of <math>\{998, 999, 1000\}</math>
+
<b>Case 5:</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.
+
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.
  
  
 
Now that we have counted all of the cases, we add them.
 
Now that we have counted all of the cases, we add them.
  
−
<math>0 + 7 + 15 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
+
<math>0 + 1 + 7 + 14 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
  
−
~dragonchomper, additional edits by [[User:emerald_block|emerald_block]]
+
~dragonchomper, additional edits by emerald_block
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:17, 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 $\frac{998}n$ 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

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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