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

(Solution 2 (Casework))
(Solution 2 (Casework))
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So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</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 we already know that <math>n</math> \neq <math>1</math>.
+
However, we have to subtract <math>1</math>, because we have counted <math>1</math> when we already know that <math>n \neq 1</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 18:50, 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 we already know that $n \neq 1$.

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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