2020 AMC 10A Problems/Problem 22

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)

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, even more edits by ericshi1685

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

Writing out $n = 1, 2, 3, 4 ... 11$ , we see that the answer cannot be more than the number of divisors of $998, 999, 1000$ since all $n$ satisfying the problem requirements are among the divisors of $998, 999, 1000$ . There are $28$ total divisors, and we subtract $3$ from the start because we count $1$ , which never works, thrice.

From the divisors of $998$ , note that $499$ and $998$ don't work. 2 to subtract. Also note that we count $2$ twice, in $998$ and $1000$ , so we have to subtract another from the running total of $25$ .

Already, we are at $22$ divisors so we conclude that the answer is $\boxed{\textbf{(A)}22}$ .

Solution 4

First, we notice the following lemma:

$\textbf{Lemma}$ : For $N, n \in \mathbb{N}$ , $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1$ if $n \mid N$ ; and $\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor$ if $n \nmid N.$

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

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

From the lemma and the given equation, we have four possible cases: $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor - 1 \qquad (1)$ $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (2)$ $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (3)$ $\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (4)$

Note that cases (2) and (3) are the cases in which the term, $\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor,$ is not divisible by $3$ . So we only need to count the number of n's for which cases (2) and (3) stand.

Case (2): By the lemma, we have $n \mid 1000$ and $n \nmid 999.$ Hence $n$ can be any factor of $1000$ except for $n = 1$ . Since $1000 = 2^3 * 5^3,$ there are $(3+1)(3+1) - 1 = 15$ possible values of $n$ for this case.

Case (3): By the lemma, we have $n \mid 999$ and $n \nmid 998.$ Hence $n$ can be any factor of $999$ except for $n = 1$ . Since $999 = 3^3 * 37^1,$ there are $(3+1)(1+1) - 1 = 7$ possible values of $n$ for this case.

So in total, we have total of $15+7=\boxed{\textbf{(A)}22}$ possible $n$ 's.

~mathboywannabe

Video Solution

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

~Snore

Education, The Study of Everything

https://youtu.be/LWAYKQQX6KI

Video Solution

https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2020 AMC 10A Problems/Problem 22

Contents

Problem

Solution 1 (Casework)

Solution 2 (Solution 1 but simpler)

Solution 3 - Solution 1 but much simpler

Solution 4

Solution 4

Video Solution

Video Solution

See Also