Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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== Video Solution == | == Video Solution == | ||
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+ | by Ismail.maths | ||
+ | https://www.youtube.com/watch?v=lt34QscjTf4&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj | ||
https://youtu.be/4vdVGYXGvzg | https://youtu.be/4vdVGYXGvzg |
Revision as of 14:44, 6 January 2023
Contents
[hide]Problem
The sum can be expressed as , where and are positive integers. What is ?
Solution 1
Note that , and therefore this sum is a telescoping sum, which is equivalent to . Our answer is .
~mathboy100
Solution 2
We add to the original expression This sum clearly telescopes, thus we end up with . Thus the original expression is equal to , and .
~not_slay (+ minor LaTeX edit ~TaeKim)
Solution 3 (Induction)
By looking for a pattern, we see that and , so we can conclude by engineer's induction that the sum in the problem is equal to , for an answer of . This can be proven with actual induction as well; we have already established base cases, so now assume that for . For we get , completing the proof. ~eibc
Solution 4
Let
Note that Therefore, the answer is
~lopkiloinm
Solution 5 (Combinatorics)
Suppose you are picking a permutation of distinct elements. Suppose that the correct order of the permutation is We want to find the probability of picking the permutation in the wrong order.
Suppose that we have picked everything to the correct order except our last elements. That is we have We want pick the next element such that it does not equal to . There are ways to choose that, so we add to the probability.
Suppose that we have picked everything to the correct order except our last elements. That is we have We want pick the next element such that it does not equal to . There are ways to choose that, so we add to the probability.
This series ends up being the probability of making a permutation in the wrong order and that is of course ~lopkiloinm
Solution 6 (Desperate)
Because the fractions get smaller, it is obvious that the answer is less than , so we can safely assume that (this can also be guessed by intuition using similar math problems). Looking at the answer choices, . Because the last term consists of (and the year is ) we can guess that . Adding them yields .
~iluvme and andy_lee
Video Solution
by Ismail.maths https://www.youtube.com/watch?v=lt34QscjTf4&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj
- Whiz
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.