2023 AIME I Problems/Problem 10
Contents
Problem
There exists a unique positive integer for which the sum is an integer strictly between and . For that unique , find .
(Note that denotes the greatest integer that is less than or equal to .)
Solution (Bounds and Decimal Part Analysis, Rigorous)
Define .
First, we bound .
We establish an upper bound of . We have
We establish a lower bound of . We have
We notice that if , then . Thus,
Because and , we must have either or .
For , we get a unique . For , there is no feasible .
Therefore, . Thus .
Next, we compute .
Let , where .
We have
Therefore,
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~minor edit by KevinChen_Yay
Solution 2
We define . Since for any real number , , we have . Now, since , we have .
Now, we can solve for in terms of . We have: So, we have , and , so we have , or . Now, is much bigger than or , and since is an integer, to satsify the inequalities, we must have , or , and .
Now, we can find . We have: Now, if , then , and if , then , and so on. Testing with , we get respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for , we get: Since , this gives , and we have .
~ genius_007
Solution 3 (Quick)
We can view the floor function in this problem as simply subtracting the remainder of (mod ) from the numerator of . For example, .
Note that the congruence of (mod ) loops every time increases by 5. Also, note that the congruence of (mod ) determines the set of congruences of for each congruence of (mod ).
For example, if (mod ), the set of remainders is for (mod ). Let the sum of these elements be . Note that for each “loop” of the numerator (mod ), each element of the set will be subtracted exactly once, meaning is subtracted once for each loop. The value of the numerator will loop times (mod ) throughout the sum, as . Then
Where . Note that since , this is an approximation for because the equation disregards the remainder (mod ) when , and so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of (*). However, we will find that this is a negligible error when it comes to the inequality , so we can proceed with this approximation to solve for .
Factoring our approximation gives
We set to make , accordingly minimizing , yielding
If increases or decreases by , then changes by which clearly breaks the inequality on . Therefore (mod ) giving the set of remainders , so and our approximation yields . However, we must subtract 2, 1, and 2 (*) giving us , giving an answer of
~Spencer Danese
Solution 4 (Calculus)
Consider the integral We hope this will give a good enough* appoximation of to find However, this integral can be easily evaluated(if you know calculus) to be Because we want this to be as close to as possible, we find that should equal Then, evaluating the sum becomes trivial. Set and Then We can evaluate to be and to be (using some basic number theory). Thus, and the answer is
- It turns out that this is quite a bad approximation, but good enough for our purposes.
Video Solution by Punxsutawney Phil
Video Solution by Steven Chen
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=49XIe2jx9zg
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.