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

Revision as of 18:05, 18 October 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.

@@ Line 31: / Line 31: @@
 ===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.
+The solution presented above does not prove why <math>F(x)</math> is found by dividing <math>x</math> by <math>3</math>. 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).
+Consider the case where <math>x = 3k</math>. We shall prove that <math>F(x) = f(x, k+1)</math>.
-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.
+For all <math>x/2 >= n > k+1, x = 2n + q</math>, where <math>0 <= q <= 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>.
-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.
+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 x = 3k + 1 and x = 3k + 2. 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 himself.
 ==See Also==
 {{AIME box|year=2013|n=II|num-b=13|num-a=15}}
 {{MAA Notice}}

2013 AIME II (Problems • Answer Key • Resources)
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