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

Revision as of 21:55, 16 November 2014

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

Easy solution without strict proof

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 $5+3\times(6+...+31)+32\times 2=1512$ , so the answer is $\boxed{512}$ .

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 $x/2 >= n > k+1, x = 2n + q$ , where $0 <= q <= n$ . This is because $x > 3k + 3 = 3n$ and $x < n$ . 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 himself.

Generalized Solution

$Lemma:$ Highest remainder when $n$ is divided by $1 <= k <= n/2$ 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 $1<= k <= n$ 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.

@@ Line 38: / Line 38: @@
 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.
+===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>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>.
+Using the lemma above:
+<math>\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}] </math>
+<math>= [(120*81/2)/3 - 2*81 + (2/3)*81]
+= 1512</math>
+So the answer is <math>\boxed{512}</math>
+Proof of Lemma: It is similar to <math>The Proof</math> stated above.
 ==See Also==
 {{AIME box|year=2013|n=II|num-b=13|num-a=15}}
 {{MAA Notice}}

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