Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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== Solution 2 (Solution 1 but simpler) == | == Solution 2 (Solution 1 but simpler) == | ||
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+ | 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>. | |
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− | + | -happykeeper, additional edits by dragonchomper, even more edits by ericshi1685 | |
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− | -happykeeper, additional edits by dragonchomper | ||
== Solution 3 - Solution 1 but much simpler == | == Solution 3 - Solution 1 but much simpler == | ||
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<math>1000=2^3\cdot 5^3 \implies (3+1)(3+1)-1 \text{*factors*} \implies 15</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. | Adding them up gives a total of <math>7+15=\boxed{\textbf{(A)}22}</math> workable <math>n</math>'s. | ||
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+ | == Solution 4 == | ||
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+ | 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. | ||
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+ | 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>. | ||
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+ | Already, we are at <math>22</math> divisors so we conclude that the answer is <math>\boxed{\textbf{(A)}22}</math>. | ||
== Solution 4 == | == Solution 4 == | ||
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~Snore | ~Snore | ||
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+ | Education, The Study of Everything | ||
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+ | https://youtu.be/LWAYKQQX6KI | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=G5UVS5aM-CY&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=4 ~ MathEx | ||
==See Also== | ==See Also== |
Revision as of 11:56, 7 November 2020
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 too works (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 4
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
https://www.youtube.com/watch?v=_Ej9nnHS07s
~Snore
Education, The Study of Everything
Video Solution
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.