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

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== Solution ==
 
== Solution ==
 
Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. If the expression 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 divisible by <math>999</math> but not <math>1000</math>, or <math>(a, a, a + 1)</math>, which would imply that <math>n</math> is divisible by <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> is divisible by 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 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 divisible by <math>999</math> but not <math>1000</math>, or <math>(a, a, a + 1)</math>, which would imply that <math>n</math> is divisible by <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> is divisible by 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>.
 +
 +
==Video Solution==
 +
https://youtu.be/Ozp3k2464u4
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 +
~IceMatrix
  
 
==See Also==
 
==See Also==

Revision as of 22:23, 31 January 2020

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

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 divisible by $999$ but not $1000$, or $(a, a, a + 1)$, which would imply that $n$ is divisible by $1000$ 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 $1$ is divisible by both $999$ and $1000$, and since $1$ is counted twice, the answer is $16 + 8 - 2 = \boxed{\textbf{(A) }22}$.

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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