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

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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, to get
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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.
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+
This gives us
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<cmath>\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor</cmath>
+
<math></math>\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor<math> = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3</math> which is divisible by 3.
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This gives us <math>\frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3</math> which is divisible by 3.
 
 
Therefore, the case <math>n = 1</math> does not work.
 
Therefore, the case <math>n = 1</math> does not work.
  

Revision as of 21:20, 1 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 $a = \left\lfloor \frac{998}n \right\rfloor$

Notice that for every integer $n \neq 1$,

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

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

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

This is due to the fact that $998$, $999$, and $1000$ share no common factors collectively (other than 1).


Note that $n = 1$ doesn't work; to prove this, we just have to substitute $1$ for $n$ in the expression. This gives us $$ (Error compiling LaTeX. Unknown error_msg)\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor$= \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3$ which is divisible by 3. Therefore, the case $n = 1$ does not work.


Now, we test the three cases mentioned above.


Case 1: $n$ divides $998$

The first case does not work, as the three terms in the expression must be $(a, a, a)$, as mentioned above, so the sum becomes $3a$, which is divisible by $3$.


Case 2: $n$ divides $999$

Because $n$ divides $999$, the number of possibilities for $n$ is the same as the number of factors of $999$, excluding $1$. $999$ = $3^3 \cdot 37^1$ So, the total number of factors of $999$ is $4 \cdot 2 = 8$.

However, we have to subtract $1$, because the case $n = 1$ doesn't work, as mentioned previously. $8 - 1 = 7$

We now do the same for the third and last case.


Case 3: $n$ divides $1000$

Because $n$ divides $1000$, the number of possibilities for $n$ is the same as the number of factors of $1000$, excluding $1$. $1000$ = $5^3 \cdot 2^3$ So, the total number of factors of $1000$ is $4 \cdot 4 = 16$.

Again, we have to subtract $1$, for the reason stated in Case 2. $16 - 1 = 15$


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

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


~dragonchomper

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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