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

Revision as of 06:09, 16 September 2022

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 $5+3\times(6+...+31)+32+32=1512$ , 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

@@ Line 52: / Line 52: @@
 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>.

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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