2013 AIME II Problems/Problem 14

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.

2013 AIME II (Problems • Answer Key • Resources)
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2013 AIME II Problems/Problem 14

Contents

Problem 14

Solution

Easy solution without strict proof

The Proof

See Also