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I need some ideas (preferable some tricks) for solving these two problems:

Find the largest number $n$ such that $(2004!)!$ is divisible by $((n!)!)!$

For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?

For the second one the suggested solution is like this : $ 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12$

But I can't understand the approach,any ideas?

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Hint for the 1st question: $x!\mid y!$ if and only if $x\le y$. Also, factorial is an increasing function. –  Jyrki Lahtonen Aug 13 '11 at 6:56
    
@Jyrki Lahtonen:That's a very precise mathematical hint:-)Thanks. –  Quixotic Aug 13 '11 at 7:04
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4 Answers

up vote 11 down vote accepted

A way to do the second problem is the following. Check small values of $n<8$ by hand (nothing there). Then assume that $n\ge8$. Now $2^8+2^{11}+2^n=2^8(1+8+2^{n-8})$ is a perfect square, iff the latter factor $9+2^{n-8}$ is a perfect square also. But $$ 9+2^{n-8}=m^2\Leftrightarrow 2^{n-8}=m^2-9=(m-3)(m+3). $$ So by the unique factorization both $m-3$ and $m+3$ must be powers of two. The difference between these two factors is 6, and the differences between powers of 2 are larger than 6 unless both powers are at most 8. The only solution is thus $m=5$, $n=12$.

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If $n=7$ then $n!=5040\gt2004$, so....

If $n=6$ then $n!=720\lt2004$, so....

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Oh, reading this made me feel silly! My first read through the question made me think of very complicated things. But this reminds me that everything in math is impossible, until it's trivial. –  mixedmath Aug 13 '11 at 6:58
    
This is exactly the same way I solved it :-) –  Quixotic Aug 13 '11 at 7:01
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The second one is true. Because $$(2^{4} + 2^{6})^{2} = 2^{8} + 2 \cdot 2^{4} \cdot 2^{6} + 2^{12}$$ so your $n=12$.

As far as I know, the main idea is to write $2^{8}+2^{11}+2^{n}$ as $(2^4)^{2} + 2 \cdot 2^{4} \cdot x + x^{2}$. Then you will have to manipulate what $x$ is and intuition say $x=2^{6}$.

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+1,the second one is so simpleeeee :-) Thanks for posting :-) –  Quixotic Aug 13 '11 at 6:49
    
@FoolFormath: Welcome. Glad I could help :) –  user9413 Aug 13 '11 at 6:50
    
I rolled back,may be somebody might post a different approach for the first one :) –  Quixotic Aug 13 '11 at 6:51
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So far as I can see, you've proved it's a square for $n=12$, but not that it's a square only for $n=12$. But maybe OP doesn't need that. –  Gerry Myerson Aug 13 '11 at 6:52
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Regarding the first question,

If $a!\leq b!$,

then $a\leq b$.

So here,

$((n!)!)!\leq (2004!)!$, implies $(n!)!\leq 2004!$,

which further implies $n!\leq 2004$,

therefore $n\leq 6$.

P.S edit: Didn't see that the solution was already posted.

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