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

Revision as of 18:24, 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 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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		+	==Problem==
	For how many positive integers <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>not divisible by <math>3</math>? (Recall that <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.)		For how many positive integers <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>not divisible by <math>3</math>? (Recall that <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.)

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