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

Revision as of 21:47, 29 March 2022

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 the fact that $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor$ to the act that $n | 999$ . 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

@@ Line 5: / Line 5: @@
 <math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math>
-== Solution 1 (Casework) ==
+== Solution 1 ==
+Clearly, <math>n=1</math> fails. Except for the special case of <math>n=1</math>,
+<cmath>\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor</cmath>
+equals either <math>0</math> or <math>1</math>. If it equals <math>0</math>, this implies that <math>\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor</math>, so their sum is clearly a multiple of <math>3</math>, so this will always fail. If it equals <math>1</math>, the sum of the three floor terms is <math>3 \left\lfloor \frac{999}{n} \right\rfloor \pm 1</math>, so it is never a multiple of <math>3</math>. Thus, we are looking for all <math>n \neq 1</math> such that
+<cmath>\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1.</cmath>
+This implies that either
+<cmath>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor,</cmath>
+or
+<cmath>\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor.</cmath>
-<b>Expression:</b>
+Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer <math>a</math> such that
-<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>
+<cmath>\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999.*</cmath>
+Analogously, the second equation implies that
+<cmath>n | 1000.</cmath>
+So our only <math>n</math> that satisfy this condition are <math>n \neq 1</math> that divide <math>999</math> or <math>1000</math>. Using the method to find the number of divisors of a number, we see that <math>999</math> has <math>8</math> divisors and <math>1000</math> has <math>16</math> divisors. Their only common factor is <math>1</math>, so there are <math>8+16-1 = 23</math> positive integers that divide either <math>999</math> or <math>1000</math>. Since the integer <math>1</math> is a special case and does not count, we must subtract this from our <math>23</math>, so our final answer is <math>23-1 = \boxed{\textbf{(C) } 22}.</math>
-<b>Solution:</b>
+*While this observation may seem strange, it is actually "trivial by intuition" to go straight from the fact that <math>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor</math> to the act that <math>n | 999</math>. In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.
-Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>
+~ihatemath123
-Since <math>\frac{1000}n - \frac{998}n = \frac{2}n</math>, for any integer <math>n \geq 2</math>, the difference between the largest and smallest terms before the <math>\lfloor x \rfloor</math> function is applied is less than or equal to <math>1</math>, and thus the terms must have a range of <math>1</math> or less after the function is applied.
+== Solution 2 ==
-This means that for every integer <math>n \geq 2</math>,
-<math>\bullet</math> if <math>\frac{998}n</math> is an integer and <math>n \neq 2</math>, then the three terms in the expression above must be <math>(a, a, a)</math>,
-<math>\bullet</math> if <math>\frac{998}n</math> is an integer because <math>n = 2</math>, then <math>\frac{1000}n</math> will be an integer and will be <math>1</math> greater than <math>\frac{998}n</math>; thus the three terms in the expression must be <math>(a, a, a + 1)</math>,
-<math>\bullet</math> if <math>\frac{999}n</math> is an integer, then the three terms in the expression above must be <math>(a, a + 1, a + 1)</math>,
-<math>\bullet</math> if <math>\frac{1000}n</math> is an integer, then the three terms in the expression above must be <math>(a, a, a + 1)</math>, and
-<math>\bullet</math> if none of <math>\left\{\frac{998}n, \frac{999}n, \frac{1000}n\right\}</math> are integral, then the three terms in the expression above must be <math>(a, a, a)</math>.
-The last statement is true because in order for the terms to be different, there must be some integer in the interval <math>\left(\frac{998}n, \frac{999}n\right)</math> or the interval <math>\left(\frac{999}n, \frac{1000}n\right)</math>. However, this means that multiplying the integer by <math>n</math> should produce a new integer between <math>998</math> and <math>999</math> or <math>999</math> and <math>1000</math>, exclusive, but because no such integers exist, the terms cannot be different, and thus, must be equal.
-<math>\bullet</math> Note that <math>n = 1</math> does not work; to prove this, we just have to substitute <math>1</math> for <math>n</math> in the expression.
-This gives us
-<math>\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</math> which is divisible by 3.
-Now, we test the five cases listed above (where <math>n \geq 2</math>)
-<b>Case 1:</b> <math>n</math> divides <math>998</math> and <math>n \neq 2</math>
-As mentioned above, the three terms in the expression are <math>(a, a, a)</math>, so the sum is <math>3a</math>, which is divisible by <math>3</math>.
-Therefore, the first case does not work (<b>0</b> cases).
-<b>Case 2:</b> <math>n</math> divides <math>998</math> and <math>n = 2</math>
-As mentioned above, in this case the terms must be <math>(a, a, a + 1)</math>, which means the sum is <math>3a + 1</math>, so the expression is not divisible by <math>3</math>. Therefore, this is <b>1</b> case that works.
-<b>Case 3:</b> <math>n</math> divides <math>999</math>
-Because <math>n</math> divides <math>999</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>999</math>.
-<math>999</math> = <math>3^3 \cdot 37^1</math>.
-So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</math>.
-However, we have to subtract <math>1</math>, because the case <math>n = 1</math> does not work, as mentioned previously. This leaves <math>8 - 1 =</math> <b>7</b> cases.
-<b>Case 4:</b> <math>n</math> divides <math>1000</math>
-Because <math>n</math> divides <math>1000</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>1000</math>.
-<math>1000</math> = <math>5^3 \cdot 2^3</math>.
-So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</math>.
-Again, we have to subtract <math>1</math>, so this leaves <math>16 - 1 = 15</math> cases.
-We have also overcounted the factor <math>2</math>, as it has been counted as a factor of <math>1000</math> and as a separate case (Case 2).
-<math>15 - 1 = 14</math>, so there are actually <b>14</b> valid cases.
-<b>Case 5:</b> <math>n</math> divides none of <math>\{998, 999, 1000\}</math>
-Similar to Case 1, the value of the terms of the expression are <math>(a, a, a)</math>. The sum is <math>3a</math>, which is divisible by 3, so this case does not work (<b>0</b> cases).
-Now that we have counted all of the cases, we add them.
-<math>0 + 1 + 7 + 14 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>.
-== Solution 2 (Solution 1 but Much Simpler) ==
-NOTE: For this problem, whenever I say <math>\text{*factors*}</math>, I will be referring to all the factors of the number except for <math>1</math>.
-Now, quickly observe that if <math>n>2</math> divides <math>998</math>, then <math>\left\lfloor {\frac{999}{n}} \right\rfloor</math> and <math>\left\lfloor {\frac{1000}{n}} \right\rfloor</math> will also round down to <math>\frac{998}{n}</math>, giving us a sum of <math>3 \cdot \frac{998}{n}</math>, which does not work for the question. However, if <math>n>2</math> divides <math>999</math>, we see that <math>\left\lfloor {\frac{998}{n}} \right\rfloor = \frac{999}{n}-1</math> and <math>\left\lfloor {\frac{1000}{n}} \right\rfloor=\left\lfloor {\frac{999}{n}} \right\rfloor</math>. This gives us a sum of <math>3 \cdot \left\lfloor {\frac{999}{n}} \right\rfloor - 1</math>, which is clearly not divisible by <math>3</math>. Using the same logic, we can deduce that <math>(n>2)|1000</math> works, too (for our problem). Thus, we need the factors of <math>999</math> and <math>1000</math> and we don't have to eliminate any because the <math>\text{gcf} (999,1000)=1</math>. But we have to be careful! See that when <math>n|998,999,1000</math>, then our problem doesn't get fulfilled. The only <math>n</math> that satisfies that is <math>n=1</math>. So, we have:
-<math>999=3^3\cdot 37 \implies (3+1)(1+1)-1 \text{*factors*} \implies 7</math>;
-<math>1000=2^3\cdot 5^3 \implies (3+1)(3+1)-1 \text{*factors*} \implies 15</math>.
-Adding them up gives a total of <math>7+15=\boxed{\textbf{(A)}22}</math> workable <math>n</math>'s.
-== Solution 3 ==
 Writing out <math>n = 1, 2, 3, 4 ... 11</math>, we see that the answer cannot be more than the number of divisors of <math>998, 999, 1000</math> since all <math>n</math> satisfying the problem requirements are among the divisors of <math>998, 999, 1000</math>. There are <math>28</math> total divisors, and we subtract <math>3</math> from the start because we count <math>1</math>, which never works, thrice.
@@ Line 98: / Line 34: @@
 Already, we are at <math>22</math> divisors so we conclude that the answer is <math>\boxed{\textbf{(A)}22}</math>.
-== Solution 4 ==
+== Solution 3 ==
 First, we notice the following lemma:

2020 AMC 10A (Problems • Answer Key • Resources)
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

Page

Toolbox

Search