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

Revision as of 10:50, 15 March 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 (Casework)

Expression: $\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$

Solution:

Let $a = \left\lfloor \frac{998}n \right\rfloor$

Since $\frac{1000}n - \frac{998}n = \frac{2}n$ , for any integer $n \geq 2$ , the difference between the largest and smallest terms before the $\lfloor x \rfloor$ function is applied is less than or equal to $1$ , and thus the terms must have a range of $1$ or less after the function is applied.

This means that for every integer $n \geq 2$ ,

$\bullet$ if $\frac{998}n$ is an integer and $n \neq 2$ , then the three terms in the expression above must be $(a, a, a)$ ,

$\bullet$ if $\frac{998}n$ is an integer because $n = 2$ , then $\frac{1000}n$ will be an integer and will be $1$ greater than $\frac{998}n$ ; thus the three terms in the expression must be $(a, a, a + 1)$ ,

$\bullet$ if $\frac{999}n$ is an integer, then the three terms in the expression above must be $(a, a + 1, a + 1)$ ,

$\bullet$ if $\frac{1000}n$ is an integer, then the three terms in the expression above must be $(a, a, a + 1)$ , and

$\bullet$ if none of $\left\{\frac{998}n, \frac{999}n, \frac{1000}n\right\}$ are integral, then the three terms in the expression above must be $(a, a, a)$ .

The last statement is true because in order for the terms to be different, there must be some integer in the interval $\left(\frac{998}n, \frac{999}n\right)$ or the interval $\left(\frac{999}n, \frac{1000}n\right)$ . However, this means that multiplying the integer by $n$ should produce a new integer between $998$ and $999$ or $999$ and $1000$ , exclusive, but because no such integers exist, the terms cannot be different, and thus, must be equal.

$\bullet$ Note that $n = 1$ does not work; to prove this, we just have to substitute $1$ for $n$ in the expression. This gives us $\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor = 998 + 999 + 1000 = 2997 = 999 \cdot 3$ which is divisible by 3.

Now, we test the five cases listed above (where $n \geq 2$ )

Case 1: $n$ divides $998$ and $n \neq 2$

As mentioned above, the three terms in the expression are $(a, a, a)$ , so the sum is $3a$ , which is divisible by $3$ . Therefore, the first case does not work (0 cases).

Case 2: $n$ divides $998$ and $n = 2$

As mentioned above, in this case the terms must be $(a, a, a + 1)$ , which means the sum is $3a + 1$ , so the expression is not divisible by $3$ . Therefore, this is 1 case that works.

Case 3: $n$ divides $999$

Because $n$ divides $999$ , the number of possibilities for $n$ is the same as the number of factors of $999$ .

$999$ = $3^3 \cdot 37^1$ . So, the total number of factors of $999$ is $4 \cdot 2 = 8$ .

However, we have to subtract $1$ , because the case $n = 1$ does not work, as mentioned previously. This leaves $8 - 1 =$ 7 cases.

Case 4: $n$ divides $1000$

Because $n$ divides $1000$ , the number of possibilities for $n$ is the same as the number of factors of $1000$ .

$1000$ = $5^3 \cdot 2^3$ . So, the total number of factors of $1000$ is $4 \cdot 4 = 16$ .

Again, we have to subtract $1$ , so this leaves $16 - 1 = 15$ cases. We have also overcounted the factor $2$ , as it has been counted as a factor of $1000$ and as a separate case (Case 2). $15 - 1 = 14$ , so there are actually 14 valid cases.

Case 5: $n$ divides none of $\{998, 999, 1000\}$

Similar to Case 1, the value of the terms of the expression are $(a, a, a)$ . The sum is $3a$ , which is divisible by 3, so this case does not work (0 cases).

Now that we have counted all of the cases, we add them.

$0 + 1 + 7 + 14 + 0 = 22$ , so the answer is $\boxed{\textbf{(A)}22}$ .

~dragonchomper, additional edits by emerald_block

Solution 2 (Solution 1 but simpler)

$*$ Note that this solution does not count a majority of cases that are important to consider in similar problems, though they are not needed for this problem, and therefore it may not work with other, similar problems.

Notice that you only need to count the number of factors of 1000 and 999, excluding 1. 1000 has 16 factors, and 999 has 8. Adding them gives you 24, but you need to subtract 2 since 1 does not work.

Therefore, the answer is 24 - 2 = $\boxed{\textbf{(A)}22}$ .

-happykeeper, additional edits by dragonchomper

Solution 3 - Solution 1 but much simpler

NOTE: For this problem, whenever I say $\text{*factors*}$ , I will be referring to all the factors of the number except for $1$ .

Now, quickly observe that if $n>2$ divides $998$ , then $\left\lfloor {\frac{999}{n}} \right\rfloor$ and $\left\lfloor {\frac{1000}{n}} \right\rfloor$ will also round down to $\frac{998}{n}$ , giving us a sum of $3 \cdot \frac{998}{n}$ , which does not work for the question. However, if $n>2$ divides $999$ , we see that $\left\lfloor {\frac{998}{n}} \right\rfloor = \frac{999}{n}-1$ and $\left\lfloor {\frac{1000}{n}} \right\rfloor=\left\lfloor {\frac{999}{n}} \right\rfloor$ . This gives us a sum of $3 \cdot \left\lfloor {\frac{999}{n}} \right\rfloor - 1$ , which is clearly not divisible by $3$ . Using the same logic, we can deduce that $(n>2)|1000$ too works (for our problem). Thus, we need the factors of $999$ and $1000$ and we don't have to eliminate any because the $\text{gcf} (999,1000)=1$ . But we have to be careful! See that when $n|998,999,1000$ , then our problem doesn't get fulfilled. The only $n$ that satisfies that is $n=1$ . So, we have: $999=3^3\cdot 37 \implies (3+1)(1+1)-1 \text{*factors*} \implies 7$ ; $1000=2^3\cdot 5^3 \implies (3+1)(3+1)-1 \text{*factors*} \implies 15$ . Adding them up gives a total of $7+15=\boxed{\textbf{(A)}22}$ workable $n$ 's.

Solution 4

First, we notice the following theorem:

If $n \mid N$ where $N, n \in \mathbb{N}$ , then $\left\lfloor \frac{N-1}{n} \right\rfloor + 1 = \left\lfloor \frac{N}{n} \right\rfloor$ . And if $n \nmid N$ , then $\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{N}{n} \right\rfloor$

Proof: Let $A = kn + r$ , with $0 \leq r < n$ . If $n \mid N$ , then $r = 0$ . Therefore $\left\lfloor \frac{N}{n} \right\rfloor = k$ and $\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1$

so $\left\lfloor \frac{N-1}{n} \right\rfloor + 1 = \left\lfloor \frac{N}{n} \right\rfloor$

If $n \nmid N$ , then $1 \leq r < n$ , therefore $\left\lfloor \frac{N}{n} \right\rfloor = k$ and $\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k$

From the theorem, we get:

$\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor$

or

$\left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor$

and

$\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor$

or

$\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor$

combined, we have:

$\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1$

or

$\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1$

Video Solution

https://www.youtube.com/watch?v=_Ej9nnHS07s

~Snore

@@ Line 108: / Line 108: @@
 If <math>n \mid N</math> where <math>N, n \in \mathbb{N}</math>, then <math>\left\lfloor \frac{N-1}{n} \right\rfloor + 1 = \left\lfloor \frac{N}{n} \right\rfloor</math>. And if <math>n \nmid N</math>, then <math>\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{N}{n} \right\rfloor</math>
-Proof: Let <math>A = kn + r</math>, with <math>0 \leq r < n</math>. If <math>n \mid N</math>, then <math>r = 0</math>. Therefore <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math> and <math>\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1</math> so <math>\left\lfloor \frac{N-1}{n} \right\rfloor + 1 = \left\lfloor \frac{N}{n} \right\rfloor</math>
+Proof: Let <math>A = kn + r</math>, with <math>0 \leq r < n</math>. If <math>n \mid N</math>, then <math>r = 0</math>. Therefore <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math> and <math>\left\lfloor \frac{N-1}{n} \right\rfloor = \left\lfloor \frac{(k-1)n+n-1}{n} \right\rfloor = k-1 + \left\lfloor \frac{n-1}{n} \right\rfloor = k-1</math>
+so <math>\left\lfloor \frac{N-1}{n} \right\rfloor + 1 = \left\lfloor \frac{N}{n} \right\rfloor</math>
+If <math>n \nmid N</math>, then <math>1 \leq r < n</math>, therefore <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math> and <math>\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k</math>
+From the theorem, we get:
+<math>\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor</math>
+or
+<math>\left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor</math>
+and
+<math>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor</math>
+or
+<math>\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor</math>
+combined, we have:
+<math>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1</math>
+or
+<math>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1</math>
 ==Video Solution==

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