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 2)
Line 8: Line 8:
 
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. If the expression <math>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</math> is not divisible by <math>3</math>, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, which would imply that <math>n</math> is a divisor of <math>999</math> but not <math>1000</math>, or <math>(a, a, a + 1)</math>, which would imply that <math>n</math> is a divisor of <math>1000</math> but not <math>999</math>. <math>999 = 3^3 \cdot 37</math> has <math>4 \cdot 2 = 8</math> factors, and <math>1000 = 2^3 \cdot 5^3</math> has <math>4 \cdot 4 = 16</math> factors. However, <math>n = 1</math> does not work because <math>1</math> a divisor of both <math>999</math> and <math>1000</math>, and since <math>1</math> is counted twice, the answer is <math>16 + 8 - 2 = \boxed{\textbf{(A) }22}</math>.
 
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. If the expression <math>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</math> is not divisible by <math>3</math>, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, which would imply that <math>n</math> is a divisor of <math>999</math> but not <math>1000</math>, or <math>(a, a, a + 1)</math>, which would imply that <math>n</math> is a divisor of <math>1000</math> but not <math>999</math>. <math>999 = 3^3 \cdot 37</math> has <math>4 \cdot 2 = 8</math> factors, and <math>1000 = 2^3 \cdot 5^3</math> has <math>4 \cdot 4 = 16</math> factors. However, <math>n = 1</math> does not work because <math>1</math> a divisor of both <math>999</math> and <math>1000</math>, and since <math>1</math> is counted twice, the answer is <math>16 + 8 - 2 = \boxed{\textbf{(A) }22}</math>.
  
−
== Solution 2 ==
+
== Solution 2 (Casework) ==
−
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that for any 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{998}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>.
+
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{998}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 998, 999, and 1000 share no factors other than 1.
−
This is due to the fact that 998, 999, and 1000 share no factors other than 1.
+
So, there are two cases:
−
 
+
<b>Case 1: <math>n</math> divides <math>999</math><\b>
 +
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>, excluding <math>1</math>.
 +
<math>999</math> = <math>3^3 \cdot 37^1</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 we have counted <math>1</math> when <math>n</math> \neq <math>1</math>.
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/Ozp3k2464u4
 
https://youtu.be/Ozp3k2464u4

Revision as of 18:46, 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

Let $a = \left\lfloor \frac{998}n \right\rfloor$. If the expression $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$ is not divisible by $3$, then the three terms in the expression must be $(a, a + 1, a + 1)$, which would imply that $n$ is a divisor of $999$ but not $1000$, or $(a, a, a + 1)$, which would imply that $n$ is a divisor of $1000$ but not $999$. $999 = 3^3 \cdot 37$ has $4 \cdot 2 = 8$ factors, and $1000 = 2^3 \cdot 5^3$ has $4 \cdot 4 = 16$ factors. However, $n = 1$ does not work because $1$ a divisor of both $999$ and $1000$, and since $1$ is counted twice, the answer is $16 + 8 - 2 = \boxed{\textbf{(A) }22}$.

Solution 2 (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 $\frac{998}n$ 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 factors other than 1. So, there are two cases: Case 1: $n$ divides $999$<\b> 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 we have counted $1$ when $n$ \neq $1$.

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=116442"