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

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== Solution 1 (Casework) ==
 
== Solution 1 (Casework) ==
 
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that for every integer <math>n \neq 1</math>, if <math>\frac{998}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a)</math>, if <math>\frac{999}n</math> is an integer, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, and if <math>\frac{1000}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a + 1)</math>. This is due to the fact that <math>998</math>, <math>999</math>, and <math>1000</math> share no common factors other than 1.
 
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that for every integer <math>n \neq 1</math>, if <math>\frac{998}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a)</math>, if <math>\frac{999}n</math> is an integer, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, and if <math>\frac{1000}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a + 1)</math>. This is due to the fact that <math>998</math>, <math>999</math>, and <math>1000</math> share no common factors other than 1.
−
Note that <math>n = 1</math> doesn't work; to prove this, all we have to do is substitute <math>1</math> for <math>n</math>. We end up with <math>\frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997  
+
Note that <math>n = 1</math> doesn't work; to prove this, all we have to do is substitute <math>1</math> for <math>n</math>. We end up with <math>\frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997</math>
  
−
= 999 \cdot 3</math> which is divisible by 3.
+
<math>= 999 \cdot 3</math> which is divisible by 3.
 
So, there are two cases:
 
So, there are two cases:
  

Revision as of 19:58, 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 1 (Casework)

Let $a = \left\lfloor \frac{998}n \right\rfloor$. Notice that for every integer $n \neq 1$, if $\frac{998}n$ is an integer, then the three terms in the expression must be $(a, a, a)$, if $\frac{999}n$ is an integer, then the three terms in the expression must be $(a, a + 1, a + 1)$, and if $\frac{1000}n$ is an integer, then the three terms in the expression must be $(a, a, a + 1)$. This is due to the fact that $998$, $999$, and $1000$ share no common factors other than 1. Note that $n = 1$ doesn't work; to prove this, all we have to do is substitute $1$ for $n$. We end up with $\frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997$

$= 999 \cdot 3$ which is divisible by 3. So, there are two cases:


Case 1: $n$ 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 second case.


Case 2: $n$ divides $1000$

$1000$ = $5^3 \cdot 2^3$

So, the total number of factors of $1000$ is $4 \cdot 4 = 16$.

Again, we have to subtract $1$, for the reason mentioned above in Case 1.

$16 - 1 = 15$


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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