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

Revision as of 18:25, 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 if $\frac{998}n$ is divisible by $3$ , then the three terms in the expression must be $(a, a, a)$ , if $\frac{998}n$ is divisible by $3$ , then the three terms in the expression must be $(a, a + 1, a + 1)$ , and if if $\frac{1000}n$ is divisible by $3$ , then the three terms in the expression must be $(a, a, a + 1)$ .

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

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.

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 == Solution 1 ==
-Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. If the expression <math>n \le 1000</math> is<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>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><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}$.
 == Solution 2 ==

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