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)

$*$ 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, 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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS