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2006 AIME I Problems/Problem 13

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Problem

For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square.

Solution

Given $g : x \mapsto \max_{j : 2^j | x} 2^j$ , consider $S_n = g(2) + \cdots + g(2^n)$ . Define $S = \{2, 4, \ldots, 2^n\}$ . There are $2^0$ elements of $S$ that are divisible by $2^n$ , $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ elements of $S$ that are divisible by $2^1$ but not by $2^2$ .

Thus $S_n = 2^0\cdot2^n + 2^0\cdot2^{n-1} + 2^1\cdot2^{n-2} + \cdots + 2^{n-2}\cdot2^1 = 2^n + (n-1)2^{n-1} = 2^{n-1}(n+1)$ , so we need $2^{2k} | n+1$ for $k \in \N$ . Now notice we also require $n < 1000$ , so if $16 | n+1$ also (but $32 \not | \, n+1$ ), then $\frac{n+1}{16} \le 62$ , so we have $n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2$ . If $16 \not | \, n+1$ , then $\frac{n+1}{4} \le 250$ , so we have $n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2$ . Finally, $n+1$ could possibly be $64, 64 \cdot 3^2$ or 256. The maximum possible $n$ is thus $4\cdot 3^2 \cdot 5^2 - 1 = 899$ .

See also

2006 AIME I (Problems • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_13"

Category: Intermediate Number Theory Problems

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