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

Revision as of 18:47, 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 $999$

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

@@ Line 11: / Line 11: @@
 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.
 So, there are two cases:
-<b>Case 1: <math>n</math> divides <math>999</math><\b>
+<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==
 https://youtu.be/Ozp3k2464u4

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

Page

Toolbox

Search