Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AIME II Problems/Problem 14"

m (→‎The Proof)
m
 
(14 intermediate revisions by 10 users not shown)
Line 4: Line 4:
 
==Solution==
 
==Solution==
  
−
===Easy solution without strict proof===
+
===The Pattern===
  
 
We can find that  
 
We can find that  
Line 28: Line 28:
 
<math>100\equiv 32 \pmod{34}</math>
 
<math>100\equiv 32 \pmod{34}</math>
  
−
So the sum is <math>5+3\times(6+...+31)+32\times 2=1512</math>, so the answer is <math>\boxed{512}</math>.
+
So the sum is <math>6+3\times(6+...+31)+31+32=1512</math>,it is also 17+20+23+...+95, so the answer is <math>\boxed{512}</math>.
 +
By: Kris17
 +
 
 +
===The Intuition===
 +
First, let's see what happens if we remove a restriction. Let's define <math>G(x)</math> as
 +
 
 +
<math>G(x):=\max_{\substack{1\le k}} f(n, k)</math>
 +
 
 +
Now, if you set <math>k</math> as any number greater than <math>n</math>, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence <math>H(x)</math> is defined as
 +
 
 +
<math>H(x):=\max_{\substack{1\le k\le n}} f(n, k)</math>
 +
 
 +
Now, after some thought, we find that if we set <math>k=\lfloor \frac{n}{2} \rfloor+1</math> we get a remainder of <math>\lfloor \frac{n-1}{2} \rfloor</math>, the max possible. Once we have gotten this far, it is easy to see that the original equation,
 +
 
 +
<math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>
 +
 
 +
has a solution with <math>k=\lfloor \frac{n}{3} \rfloor+1</math>.
 +
 
 +
<math>W^5</math>~Rowechen
  
 
===The Proof===
 
===The Proof===
Line 34: Line 52:
  
 
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\geq n > k+1, x = 2n + q</math>, where <math>0\leq q\leq n</math>. This is because <math>x > 3k + 3 = 3n</math> and <math>x < n</math>. Also, as n 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>.
+
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>.
  
−
Similar proofs apply for <math>x = 3k + 1</math> and <math>x = 3k + 2</math>. The reader should feel free to derive these proofs himself.
+
Similar proofs apply for <math>x = 3k + 1</math> and <math>x = 3k + 2</math>. The reader should feel free to derive these proofs themself.
  
 
===Generalized Solution===
 
===Generalized Solution===
  
−
<math>Lemma:</math> Highest remainder when <math>n</math> is divided by <math>1 <= k <= n/2</math> is obtained for <math>k_0 = (n + (3 - n \pmod{3}))/3</math> and the remainder thus obtained is <math>(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]</math>.  
+
<math>Lemma:</math> Highest remainder when <math>n</math> is divided by <math>1\leq k\leq n/2</math> is obtained for <math>k_0 = (n + (3 - n \pmod{3}))/3</math> and the remainder thus obtained is <math>(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]</math>.  
  
−
<math>Note:</math> This is the second highest remainder when <math>n</math> is divided by <math>1<= k <= n</math> and the highest remainder occurs when <math>n</math> is divided by <math>k_M</math> = <math>(n+1)/2</math> for odd <math>n</math> and  <math>k_M</math> = <math>(n+2)/2</math> for even <math>n</math>.
+
<math>Note:</math> This is the second highest remainder when <math>n</math> is divided by <math>1\leq k\leq n</math> and the highest remainder occurs when <math>n</math> is divided by <math>k_M</math> = <math>(n+1)/2</math> for odd <math>n</math> and  <math>k_M</math> = <math>(n+2)/2</math> for even <math>n</math>.
  
 
Using the lemma above:
 
Using the lemma above:
Line 56: Line 74:
  
 
Kris17
 
Kris17
 +
 +
 +
==Video Solution==
 +
 +
https://youtu.be/mQ4_1dp8Wm8?si=Ae5HAc0cZQAjdtWl
 +
 +
~MathProblemSolvingSkills.com
 +
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2013|n=II|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:41, 26 June 2024

Problem 14

For positive integers $n$ and $k$, let $f(n, k)$ be the remainder when $n$ is divided by $k$, and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$. Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$.

Solution

The Pattern

We can find that

$20\equiv 6 \pmod{7}$

$21\equiv 5 \pmod{8}$

$22\equiv 6 \pmod{8}$

$23\equiv 7 \pmod{8}$

$24\equiv 6 \pmod{9}$

$25\equiv 7 \pmod{9}$

$26\equiv 8 \pmod{9}$

Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get

$99\equiv 31 \pmod{34}$

$100\equiv 32 \pmod{34}$

So the sum is $6+3\times(6+...+31)+31+32=1512$,it is also 17+20+23+...+95, so the answer is $\boxed{512}$. By: Kris17

The Intuition

First, let's see what happens if we remove a restriction. Let's define $G(x)$ as

$G(x):=\max_{\substack{1\le k}} f(n, k)$

Now, if you set $k$ as any number greater than $n$, you get n, obviously the maximum possible. That's too much freedom; let's restrict it a bit. Hence $H(x)$ is defined as

$H(x):=\max_{\substack{1\le k\le n}} f(n, k)$

Now, after some thought, we find that if we set $k=\lfloor \frac{n}{2} \rfloor+1$ we get a remainder of $\lfloor \frac{n-1}{2} \rfloor$, the max possible. Once we have gotten this far, it is easy to see that the original equation,

$F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$

has a solution with $k=\lfloor \frac{n}{3} \rfloor+1$.

$W^5$~Rowechen

The Proof

The solution presented above does not prove why $F(x)$ is found by dividing $x$ by $3$. Indeed, that is the case, as rigorously shown below.

Consider the case where $x = 3k$. We shall prove that $F(x) = f(x, k+1)$. For all $x/2\ge n > k+1, x = 2n + q$, where $0\le q< n$. This is because $x < 3k + 3 < 3n$ and $x \ge 2n$. Also, as $n$ increases, $q$ decreases. Thus, $q = f(x, n) < f(x, k+1) = k - 2$ for all $n > k+1$. Consider all $n < k+1. f(x, k) = 0$ and $f(x, k-1) = 3$. Also, $0 < f(x, k-2) < k-2$. Thus, for $k > 5, f(x, k+1) > f(x, n)$ for $n < k+1$.

Similar proofs apply for $x = 3k + 1$ and $x = 3k + 2$. The reader should feel free to derive these proofs themself.

Generalized Solution

$Lemma:$ Highest remainder when $n$ is divided by $1\leq k\leq n/2$ is obtained for $k_0 = (n + (3 - n \pmod{3}))/3$ and the remainder thus obtained is $(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]$.

$Note:$ This is the second highest remainder when $n$ is divided by $1\leq k\leq n$ and the highest remainder occurs when $n$ is divided by $k_M$ = $(n+1)/2$ for odd $n$ and $k_M$ = $(n+2)/2$ for even $n$.

Using the lemma above:

$\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}]$ $= [(120*81/2)/3 - 2*81 + (2/3)*81] = 1512$

So the answer is $\boxed{512}$

Proof of Lemma: It is similar to $The Proof$ stated above.

Kris17


Video Solution

https://youtu.be/mQ4_1dp8Wm8?si=Ae5HAc0cZQAjdtWl

~MathProblemSolvingSkills.com


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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_14&oldid=221142"