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2020 AMC 10A Problems/Problem 22

Revision as of 18:35, 1 February 2020 by Dragonchomper (talk | contribs) (→‎Solution 2)

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 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.

Video Solution

https://youtu.be/Ozp3k2464u4

~IceMatrix

See Also

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

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