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Difference between revisions of "2006 AIME I Problems/Problem 13"
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== Problem ==
 
== Problem ==
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For each even positive integer <math> x, </math> let <math> g(x) </math> denote the greatest power of 2 that divides <math> x. </math> For example, <math> g(20)=4 </math> and <math> g(16)=16. </math> For each positive integer <math> n, </math> let <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a perfect square.
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For each [[even integer | even]] [[positive integer]] <math> x </math>, let <math> g(x) </math> denote the greatest power of 2 that [[divisor | divides]] <math> x. </math> For example, <math> g(20)=4 </math> and <math> g(16)=16. </math> For each positive integer <math> n, </math> let <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]].
  
  
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== Solution 1 ==
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Given <math>g : x \mapsto \max_{j : 2^j | x} 2^j</math>, consider <math>S_n = g(2) + \cdots + g(2^n)</math>. Define <math>S = \{2, 4, \ldots, 2^n\}</math>. There are <math>2^0</math> elements of <math>S</math> that are divisible by <math>2^n</math>, <math>2^1 - 2^0 = 2^0</math> elements of <math>S</math> that are divisible by <math>2^{n-1}</math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math> but not by <math>2^2</math>.
  
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Thus
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<cmath>\begin{align*}
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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).\end{align*}</cmath> Let <math>2^k</math> be the highest power of <math>2</math> that divides <math>n+1</math>. Thus by the above formula, the highest power of <math>2</math> that divides <math>S_n</math> is <math>2^{k+n-1}</math>. For <math>S_n</math> to be a perfect square, <math>k+n-1</math> must be even. If <math>k</math> is odd, then <math>n+1</math> is even, hence <math>k+n-1</math> is odd, and <math>S_n</math> cannot be a perfect square. Hence <math>k</math> must be even. In particular, as <math>n<1000</math>, we have five choices for <math>k</math>, namely <math>k=0,2,4,6,8</math>.
  
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If <math>k=0</math>, then <math>n+1</math> is odd, so <math>k+n-1</math> is odd, hence the largest power of <math>2</math> dividing <math>S_n</math> has an odd exponent, so <math>S_n</math> is not a perfect square.
  
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== Solution ==
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In the other cases, note that <math>k+n-1</math> is even, so the highest power of <math>2</math> dividing <math>S_n</math> will be a perfect square. In particular, <math>S_n</math> will be a perfect square if and only if <math>(n+1)/2^{k}</math> is an odd perfect square.
  
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If <math>k=2</math>, then <math>n<1000</math> implies that <math>\frac{n+1}{4} \le 250</math>, so we have <math>n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2</math>.
  
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If <math>k=4</math>, then <math>n<1000</math> implies that <math>\frac{n+1}{16} \le 62</math>, so <math>n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2</math>.
  
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If <math>k=6</math>, then <math>n<1000</math> implies that <math>\frac{n+1}{64}\le 15</math>, so <math>n+1=64,64\cdot 3^2</math>.
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If <math>k=8</math>, then <math>n<1000</math> implies that <math>\frac{n+1}{256}\le 3</math>, so <math>n+1=256</math>.
 +
 +
Comparing the largest term in each case, we find that the maximum possible <math>n</math> such that <math>S_n</math> is a perfect square is <math>4\cdot 3^2 \cdot 5^2 - 1 = \boxed{899}</math>.
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== Solution 2 ==
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First note that <math>g(k)=1</math> if <math>k</math> is odd and <math>2g(k/2)</math> if <math>k</math> is even.
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so <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).</math>
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<math>2k-1</math> must be odd so this reduces to <math>2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).</math>  Thus <math>S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.</math> Further noting that <math>S_0=1</math> we can see that <math>S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).  </math> which is the same as above.  To simplify the process of finding the largest square <math>S_n</math> we can note that if <math>n-1</math> is odd then <math>n+1</math> must be exactly divisible by an odd power of <math>2</math>.  However, this means <math>n+1</math> is even but it cannot be.  Thus <math>n-1</math> is even and <math>n+1</math> is a large even square.  The largest even square <math>< 1000</math> is <math>900</math> so <math>n+1= 900 => n= \boxed{899}</math>
  
 
== See also ==
 
== See also ==
−
* [[2006 AIME I Problems]]
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{{AIME box|year=2006|n=I|num-b=12|num-a=14}}
 +
 
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 14:34, 16 January 2017

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 1

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 \begin{align*} 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).\end{align*} Let $2^k$ be the highest power of $2$ that divides $n+1$. Thus by the above formula, the highest power of $2$ that divides $S_n$ is $2^{k+n-1}$. For $S_n$ to be a perfect square, $k+n-1$ must be even. If $k$ is odd, then $n+1$ is even, hence $k+n-1$ is odd, and $S_n$ cannot be a perfect square. Hence $k$ must be even. In particular, as $n<1000$, we have five choices for $k$, namely $k=0,2,4,6,8$.

If $k=0$, then $n+1$ is odd, so $k+n-1$ is odd, hence the largest power of $2$ dividing $S_n$ has an odd exponent, so $S_n$ is not a perfect square.

In the other cases, note that $k+n-1$ is even, so the highest power of $2$ dividing $S_n$ will be a perfect square. In particular, $S_n$ will be a perfect square if and only if $(n+1)/2^{k}$ is an odd perfect square.

If $k=2$, then $n<1000$ implies that $\frac{n+1}{4} \le 250$, so we have $n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2$.

If $k=4$, then $n<1000$ implies that $\frac{n+1}{16} \le 62$, so $n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2$.

If $k=6$, then $n<1000$ implies that $\frac{n+1}{64}\le 15$, so $n+1=64,64\cdot 3^2$.

If $k=8$, then $n<1000$ implies that $\frac{n+1}{256}\le 3$, so $n+1=256$.

Comparing the largest term in each case, we find that the maximum possible $n$ such that $S_n$ is a perfect square is $4\cdot 3^2 \cdot 5^2 - 1 = \boxed{899}$.

Solution 2

First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even. so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$. However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$

See also

2006 AIME I (Problems • Answer Key • Resources)
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Problem 14
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