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

Revision as of 19:03, 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 $\frac{999}n$ 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 the case $n = 1$ when we already know that $n \neq 1$ .

$8 - 1 = 7$

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

@@ Line 7: / Line 7: @@
 == 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{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 <math>998</math>, <math>999</math>, and <math>1000</math> share no 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{999}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 <math>998</math>, <math>999</math>, and <math>1000</math> share no factors other than 1.
 So, there are two cases:

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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Difference between revisions of "2020 AMC 10A Problems/Problem 22"

Revision as of 19:03, 1 February 2020

Contents

Problem

Solution 1 (Casework)

Video Solution

See Also