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

Clearly, $n=1$ fails. Except for the special case of $n=1$ , $\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor$ equals either $0$ or $1$ . If it equals $0$ , this implies that $\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor$ , so their sum is clearly a multiple of $3$ , so this will always fail. If it equals $1$ , the sum of the three floor terms is $3 \left\lfloor \frac{999}{n} \right\rfloor \pm 1$ , so it is never a multiple of $3$ . Thus, we are looking for all $n \neq 1$ such that $\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1.$ This implies that either $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor,$ or $\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor.$

Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer $a$ such that $\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999.*$ Analogously, the second equation implies that $n | 1000.$ So our only $n$ that satisfy this condition are $n \neq 1$ that divide $999$ or $1000$ . Using the method to find the number of divisors of a number, we see that $999$ has $8$ divisors and $1000$ has $16$ divisors. Their only common factor is $1$ , so there are $8+16-1 = 23$ positive integers that divide either $999$ or $1000$ . Since the integer $1$ is a special case and does not count, we must subtract this from our $23$ , so our final answer is $23-1 = \boxed{\textbf{(C) } 22}.$

*While this observation may seem strange, it is actually "trivial by intuition" to go straight from  to . In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.

~ihatemath123

Solution 2

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 3

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 Solutions

Video Solution 1 (Simple)

Education, The Study of Everything

https://youtu.be/LWAYKQQX6KI

Video Solution 2

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

~Snore

Video Solution 3

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

Video Solution 4 (Richard Rusczyk)

https://artofproblemsolving.com/videos/amc/2020amc10a/517

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