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

Revision as of 18:35, 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

Let $a = \left\lfloor \frac{998}n \right\rfloor$ . Notice that for any $n \neq 1$ , 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.

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

@@ Line 9: / Line 9: @@
 == Solution 2 ==
-Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that 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> (unless n = 1).
+Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that for any <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.

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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