Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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+ | ==Problem== | ||
For how many positive integers <math>n \le 1000</math> is<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>not divisible by <math>3</math>? (Recall that <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.) | For how many positive integers <math>n \le 1000</math> is<cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath>not divisible by <math>3</math>? (Recall that <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.) | ||
<math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math> | <math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math> | ||
+ | |||
+ | == Solution 1 (Casework) == | ||
+ | |||
+ | <b>Expression:</b> | ||
+ | <cmath>\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor</cmath> | ||
+ | |||
+ | <b>Solution:</b> | ||
+ | |||
+ | Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~dragonchomper, additional edits by emerald_block | ||
+ | |||
+ | == Solution 2 (Solution 1 but simpler) == | ||
+ | |||
+ | Notice that you only need to count the number of factors of <math>1000</math> and <math>999</math>, excluding <math>1</math>. | ||
+ | <math>1000</math> has <math>16</math> factors, and <math>999</math> has <math>8</math>. | ||
+ | Adding them gives you <math>24</math>, but you need to subtract <math>2</math> since <math>1</math> does not work. | ||
+ | |||
+ | Therefore, the answer is <math>24 - 2 = \boxed{\textbf{(A)}22}</math>. | ||
+ | |||
+ | -happykeeper, additional edits by dragonchomper, even more edits by ericshi1685 | ||
+ | |||
+ | == Solution 3 - 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 4 == | ||
+ | |||
+ | 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. | ||
+ | |||
+ | From the divisors of <math>998</math>, note that <math>499</math> and <math>998</math> don't work. 2 to subtract. | ||
+ | Also note that we count <math>2</math> twice, in <math>998</math> and <math>1000</math>, so we have to subtract another from the running total of <math>25</math>. | ||
+ | |||
+ | Already, we are at <math>22</math> divisors so we conclude that the answer is <math>\boxed{\textbf{(A)}22}</math>. | ||
+ | |||
+ | == Solution 5 == | ||
+ | First, we notice the following lemma: | ||
+ | |||
+ | <math>\textbf{Lemma}</math>: For <math>N, n \in \mathbb{N}</math>, <math> \left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1 </math> if <math>n \mid N</math>; and <math>\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor</math> if <math>n \nmid N.</math> | ||
+ | |||
+ | <math>\textbf{Proof}</math>: Let <math>A = kn + r</math>, with <math>0 \leq r < n</math>. If <math>n \mid N</math>, then <math>r = 0</math>. Hence <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math>, <math>\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</math>, and <math>\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor + 1.</math> | ||
+ | |||
+ | If <math>n \nmid N</math>, then <math>1 \leq r < n</math>. Hence <math>\left\lfloor \frac{N}{n} \right\rfloor = k</math>, <math>\left\lfloor \frac{N-1}{n} \right\rfloor = k + \left\lfloor \frac{r-1}{n} \right\rfloor = k</math>, and <math>\left\lfloor \frac{N}{n} \right\rfloor = \left\lfloor \frac{N-1}{n} \right\rfloor.</math> | ||
+ | |||
+ | From the lemma and the given equation, we have four possible cases: | ||
+ | <cmath>\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)</cmath> | ||
+ | <cmath>\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)</cmath> | ||
+ | <cmath>\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (3)</cmath> | ||
+ | <cmath>\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor \qquad (4)</cmath> | ||
+ | |||
+ | Note that cases (2) and (3) are the cases in which the term, <math>\left\lfloor \frac{998}{n} \right\rfloor + \left\lfloor \frac{999}{n} \right\rfloor + \left\lfloor \frac{1000}{n} \right\rfloor,</math> is not divisible by <math>3</math>. 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 <math>n \mid 1000</math> and <math>n \nmid 999.</math> Hence <math>n</math> can be any factor of <math>1000</math> except for <math>n = 1</math>. Since <math>1000 = 2^3 * 5^3,</math> there are <math>(3+1)(3+1) - 1 = 15</math> possible values of <math>n</math> for this case. | ||
+ | |||
+ | Case (3): By the lemma, we have <math>n \mid 999</math> and <math>n \nmid 998.</math> Hence <math>n</math> can be any factor of <math>999</math> except for <math>n = 1</math>. Since <math>999 = 3^3 * 37^1,</math> there are <math>(3+1)(1+1) - 1 = 7</math> possible values of <math>n</math> for this case. | ||
+ | |||
+ | So in total, we have total of <math>15+7=\boxed{\textbf{(A)}22}</math> possible <math>n</math>'s. | ||
+ | |||
+ | ~mathboywannabe | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://www.youtube.com/watch?v=_Ej9nnHS07s | ||
+ | |||
+ | ~Snore | ||
+ | ==Video Solution 2== | ||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/LWAYKQQX6KI | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx | ||
==See Also== | ==See Also== |
Latest revision as of 12:43, 9 January 2021
Contents
Problem
For how many positive integers isnot divisible by ? (Recall that is the greatest integer less than or equal to .)
Solution 1 (Casework)
Expression:
Solution:
Let
Since , for any integer , the difference between the largest and smallest terms before the function is applied is less than or equal to , and thus the terms must have a range of or less after the function is applied.
This means that for every integer ,
if is an integer and , then the three terms in the expression above must be ,
if is an integer because , then will be an integer and will be greater than ; thus the three terms in the expression must be ,
if is an integer, then the three terms in the expression above must be ,
if is an integer, then the three terms in the expression above must be , and
if none of are integral, then the three terms in the expression above must be .
The last statement is true because in order for the terms to be different, there must be some integer in the interval or the interval . However, this means that multiplying the integer by should produce a new integer between and or and , exclusive, but because no such integers exist, the terms cannot be different, and thus, must be equal.
Note that does not work; to prove this, we just have to substitute for in the expression.
This gives us
which is divisible by 3.
Now, we test the five cases listed above (where )
Case 1: divides and
As mentioned above, the three terms in the expression are , so the sum is , which is divisible by . Therefore, the first case does not work (0 cases).
Case 2: divides and
As mentioned above, in this case the terms must be , which means the sum is , so the expression is not divisible by . Therefore, this is 1 case that works.
Case 3: divides
Because divides , the number of possibilities for is the same as the number of factors of .
= . So, the total number of factors of is .
However, we have to subtract , because the case does not work, as mentioned previously. This leaves 7 cases.
Case 4: divides
Because divides , the number of possibilities for is the same as the number of factors of .
= . So, the total number of factors of is .
Again, we have to subtract , so this leaves cases. We have also overcounted the factor , as it has been counted as a factor of and as a separate case (Case 2). , so there are actually 14 valid cases.
Case 5: divides none of
Similar to Case 1, the value of the terms of the expression are . The sum is , 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.
, so the answer is .
~dragonchomper, additional edits by emerald_block
Solution 2 (Solution 1 but simpler)
Notice that you only need to count the number of factors of and , excluding . has factors, and has . Adding them gives you , but you need to subtract since does not work.
Therefore, the answer is .
-happykeeper, additional edits by dragonchomper, even more edits by ericshi1685
Solution 3 - Solution 1 but much simpler
NOTE: For this problem, whenever I say , I will be referring to all the factors of the number except for .
Now, quickly observe that if divides , then and will also round down to , giving us a sum of , which does not work for the question. However, if divides , we see that and . This gives us a sum of , which is clearly not divisible by . Using the same logic, we can deduce that works, too (for our problem). Thus, we need the factors of and and we don't have to eliminate any because the . But we have to be careful! See that when , then our problem doesn't get fulfilled. The only that satisfies that is . So, we have: ; . Adding them up gives a total of workable 's.
Solution 4
Writing out , we see that the answer cannot be more than the number of divisors of since all satisfying the problem requirements are among the divisors of . There are total divisors, and we subtract from the start because we count , which never works, thrice.
From the divisors of , note that and don't work. 2 to subtract. Also note that we count twice, in and , so we have to subtract another from the running total of .
Already, we are at divisors so we conclude that the answer is .
Solution 5
First, we notice the following lemma:
: For , if ; and if
: Let , with . If , then . Hence , , and
If , then . Hence , , and
From the lemma and the given equation, we have four possible cases:
Note that cases (2) and (3) are the cases in which the term, is not divisible by . 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 and Hence can be any factor of except for . Since there are possible values of for this case.
Case (3): By the lemma, we have and Hence can be any factor of except for . Since there are possible values of for this case.
So in total, we have total of possible 's.
~mathboywannabe
Video Solution 1
https://www.youtube.com/watch?v=_Ej9nnHS07s
~Snore
Video Solution 2
Education, The Study of Everything
Video Solution 3
https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx
See Also
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 |
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