Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math> | 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. | |
− | <math>\bullet</math> if <math>\frac{ | + | This means that for every integer <math>n \geq 2</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>, | ||
<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{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> | + | <math>\bullet</math> if <math>\frac{998}n</math> is an integer and <math>n \neq 2</math> (since if <math>n = 2</math>, <math>\frac{1000}n</math> will be an integer, and it will be <math>1</math> greater than <math>\frac{998}n</math>), then the three terms in the expression above must be <math>(a, a, a)</math>, and |
− | <math>\bullet</math> if none of | + | <math>\bullet</math> if none of <math>\{\frac{998}n, \frac{999}n, \frac{1000}n\}</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>(\frac{998}n, \frac{999}n)</math> or the interval <math>(\frac{999}n, \frac{1000}n)</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 original integer cannot exist, and thus, the terms must be equal. | |
− | Note that <math>n = 1</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 | 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 = | + | <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. |
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<b>Case 1:</b> <math>n</math> divides <math>998</math> | <b>Case 1:</b> <math>n</math> divides <math>998</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. | Therefore, the first case does not work. | ||
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So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</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> | + | 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 = 7</math> cases. |
− | <math>8 - 1 = 7</math> | ||
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So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</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>, | + | Again, we have to subtract <math>1</math>, so this leaves <math>16 - 1 = 15</math> cases. |
− | <math>16 - 1 = 15</math> | ||
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− | + | <b>Case 4:</b> <math>n</math> divides none of <math>\{998, 999, 1000\}</math> | |
+ | As in 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. | ||
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<math>0 + 7 + 15 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>. | <math>0 + 7 + 15 + 0 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>. | ||
− | ~dragonchomper | + | ~dragonchomper, additional edits by [[User:emerald_block|emerald_block]] |
==Video Solution== | ==Video Solution== |
Revision as of 01:35, 2 February 2020
Problem
For how many positive integers isnot divisible by ? (Recall that is the greatest integer less than or equal to .)
Solution (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, then the three terms in the expression above must be ,
if is an integer, then the three terms in the expression above must be ,
if is an integer and (since if , will be an integer, and it will be greater than ), 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 original integer cannot exist, and thus, the terms 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 four cases listed above.
Case 1: divides
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.
Case 2: 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 cases.
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 .
Again, we have to subtract , so this leaves cases.
Case 4: divides none of
As in 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.
Now that we have counted all of the cases, we add them.
, so the answer is .
~dragonchomper, additional edits by emerald_block
Video Solution
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.