Difference between revisions of "2013 AIME II Problems/Problem 14"
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For positive integers <math>n</math> and <math>k</math>, let <math>f(n, k)</math> be the remainder when <math>n</math> is divided by <math>k</math>, and for <math>n > 1</math> let <math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>. Find the remainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>. | For positive integers <math>n</math> and <math>k</math>, let <math>f(n, k)</math> be the remainder when <math>n</math> is divided by <math>k</math>, and for <math>n > 1</math> let <math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>. Find the remainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>. | ||
− | == | + | ==Solution== |
===The Pattern=== | ===The Pattern=== | ||
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<math>100\equiv 32 \pmod{34}</math> | <math>100\equiv 32 \pmod{34}</math> | ||
− | So the sum is <math> | + | So the sum is <math>5+3\times(6+...+31)+32+32=1512</math>, so the answer is <math>\boxed{512}</math>. |
By: Kris17 | By: Kris17 | ||
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Consider the case where <math>x = 3k</math>. We shall prove that <math>F(x) = f(x, k+1)</math>. | Consider the case where <math>x = 3k</math>. We shall prove that <math>F(x) = f(x, k+1)</math>. | ||
− | For all <math>x/2\ | + | For all <math>x/2\ge n > k+1, x = 2n + q</math>, where <math>0\le q< n</math>. This is because <math>x < 3k + 3 < 3n</math> and <math>x \ge 2n</math>. Also, as <math>n</math> increases, <math>q</math> decreases. Thus, <math>q = f(x, n) < f(x, k+1) = k - 2</math> for all <math>n > k+1</math>. |
Consider all <math>n < k+1. f(x, k) = 0</math> and <math>f(x, k-1) = 3</math>. Also, <math>0 < f(x, k-2) < k-2</math>. Thus, for <math>k > 5, f(x, k+1) > f(x, n)</math> for <math>n < k+1</math>. | Consider all <math>n < k+1. f(x, k) = 0</math> and <math>f(x, k-1) = 3</math>. Also, <math>0 < f(x, k-2) < k-2</math>. Thus, for <math>k > 5, f(x, k+1) > f(x, n)</math> for <math>n < k+1</math>. | ||
Latest revision as of 06:09, 16 September 2022
Contents
Problem 14
For positive integers and , let be the remainder when is divided by , and for let . Find the remainder when is divided by .
Solution
The Pattern
We can find that
Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get
So the sum is , so the answer is . By: Kris17
The Intuition
First, let's see what happens if we remove a restriction. Let's define as
Now, if you set as any number greater than , you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence is defined as
Now, after some thought, we find that if we set we get a remainder of , the max possible. Once we have gotten this far, it is easy to see that the original equation,
has a solution with .
~Rowechen
The Proof
The solution presented above does not prove why is found by dividing by . Indeed, that is the case, as rigorously shown below.
Consider the case where . We shall prove that . For all , where . This is because and . Also, as increases, decreases. Thus, for all . Consider all and . Also, . Thus, for for .
Similar proofs apply for and . The reader should feel free to derive these proofs themself.
Generalized Solution
Highest remainder when is divided by is obtained for and the remainder thus obtained is .
This is the second highest remainder when is divided by and the highest remainder occurs when is divided by = for odd and = for even .
Using the lemma above:
So the answer is
Proof of Lemma: It is similar to stated above.
Kris17
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.