Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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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>. | Notice that you only need to count the number of factors of <math>1000</math> and <math>999</math>, excluding <math>1</math>. |
Latest revision as of 21:01, 25 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, read Solution 1 first)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.