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

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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