Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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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> | 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> | ||
− | *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. | + | *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. |
~ihatemath123 | ~ihatemath123 |
Revision as of 20:48, 29 March 2022
Contents
Problem
For how many positive integers isnot divisible by ? (Recall that is the greatest integer less than or equal to .)
Solution 1
Clearly, fails. Except for the special case of , equals either or . If it equals , this implies that , so their sum is clearly a multiple of , so this will always fail. If it equals , the sum of the three floor terms is , so it is never a multiple of . Thus, we are looking for all such that This implies that either or
Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer such that Analogously, the second equation implies that So our only that satisfy this condition are that divide or . Using the method to find the number of divisors of a number, we see that has divisors and has divisors. Their only common factor is , so there are positive integers that divide either or . Since the integer is a special case and does not count, we must subtract this from our , so our final answer is
*While this observation may seem strange, it is actually "trivial by intuition" to go straight from the fact that to the act that . In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.
~ihatemath123
Solution 2
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. 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 3
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 '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 Solutions
Video Solution 1 (Simple)
Education, The Study of Everything
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
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.