Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution (Casework))
(→‎Solution (Casework))
Line 19: Line 19:
 
<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>,
 
<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>,
  
−
<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>, and
+
<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>.
+
<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</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</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>.

Revision as of 22:13, 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 $(a, a + 1, a + 1)$,

$\bullet$ 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 $(a, a, a)$.

This is due to the fact that $998$, $999$, and $1000$ do not collectively share any common factors (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 $\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3$ which is divisible by 3. Therefore, the case $n = 1$ does not work.


Now, we test the four cases listed above.


Case 1: $n$ divides $998$

The three terms in the expression must be $(a, a, a)$, as mentioned above, 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 $n = 1$ doesn't work, as mentioned previously. $8 - 1 = 7$

We now do the same for the third 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 $1000$. $1000$ = $5^3 \cdot 2^3$ 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. $16 - 1 = 15$

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

If $n$ does not divide {$998, 999, 1000$}, then the value of the terms of the above expression are $(a, a, a)$. See if you can figure out why, although the explanation is similar to that of Case 1.

Therefore, the value of the expression 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_22&oldid=116534"