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Find maximum $n$ satisfying $(2^n)|A$. Where $A=63.64.65.66.67.68 .\cdots. 130$ .

Hung Nguyen
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  • Those periods should be $\cdot$s or equivalently $\times$s for multiplication I presume. Any ideas? – anon Sep 11 '13 at 00:01
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    Where did your first attempt fail? – Michael Albanese Sep 11 '13 at 00:01
  • In response to the deleted answer: the base-$p$ logarithm and highest power of $p$ dividing an integer can be arbitrarily far apart. – anon Sep 11 '13 at 00:16
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    There is a fairly explicit formula, by Legendre, for the highest power of $2$ dividing $n!$ for some given $n.$ So, do $130!$ and $62!$ and subtract. See http://en.wikipedia.org/wiki/Factorial#Number_theory OR http://www.cut-the-knot.org/blue/LegendresTheorem.shtml – Will Jagy Sep 11 '13 at 00:19
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    There is a fancy but computationally easy way. But you can do it the hard way. The number $64$ contributes $6$ $2$'s, the number $66$ contributes $1$, the number $68$ contributes $2$, $70$ gives $1$, $72$ gives $3$, and so on. Add up. – André Nicolas Sep 11 '13 at 00:19
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    A few proofs of Legendre's formula at http://math.stackexchange.com/questions/110553/understanding-the-proof-of-a-formula-for-pe-vert-n . I proved it by induction, you don't often see that. – Will Jagy Sep 11 '13 at 00:25
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    @WillJagy The appropriate phrase here is "Like a boss." – Pedro Sep 11 '13 at 01:45
  • @PeterTamaroff or http://www.youtube.com/watch?v=s__rX_WL100 – Will Jagy Sep 11 '13 at 01:56
  • @WillJagy You mean http://www.youtube.com/watch?v=NisCkxU544c – Pedro Sep 11 '13 at 02:05
  • @PeterTamaroff, actually the Madonna song is pretty stupid. Never really listened to it before. This, however, is very nice. The comment in the text is correct, Paillard recorded two versions, and I think this is the one I prefer; I will continue checking. What I remember in particular is the plucked string instrument... http://www.youtube.com/watch?v=aQxVE6d3Yj8 – Will Jagy Sep 11 '13 at 02:16
  • @PeterTamaroff, the Like a Boss song is quite good. Never heard of it before, but then, I wouldn't. – Will Jagy Sep 11 '13 at 02:27

2 Answers2

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Take your number, and cross out odds. You get $$130\cdot 128\cdot\;\cdots \;\cdot 68\cdot 66\cdot 64$$

Now, divide through $2$, to get $$65\cdot 64\cdot\; \cdots\; \cdot 34\cdot 33\cdot 32$$

In this step we collected $65-32+1=34$ that is $2^{34}$. Cross out odds, $$64\cdot 62\cdot 60\cdot \;\cdots\; \cdot 36\cdot 34\cdot 32$$ divide, $$32\cdot 31\cdot\;\cdots \;\cdot 16$$ and we collected $32-16+1=17$, that is $2^{17}$. Cross out odds, divide, to get $$16\cdot 15\cdots\;\cdots\;\cdot 8$$ so we collected $16-8+1=9$, so $2^9$. Finally $8\cdot 7\cdot 6 \cdot 5\cdot 4$, collected $2^5$, $4\cdot 3\cdot 2$, collected $2^3$, and $2^2$; $2$. The final count is then $$2^{34+17+9+5+3+2+1}=2^{71}$$

Wolfram approved

Pedro
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Writing your product this way should help: $$ (2^6-1)(2^6)(2^6+1)\times \dots \times (2^7)(2^7+1)(2^7+2) $$ $\Longrightarrow$ \begin{align*} 6+1+2+1+3+1+2+1+4+1+2+&\\1+3+1+2+1+5+1+2+1+3+1+&\\2+1+4+1+2+1+3+1+2+1+7+&1 =71\\ \end{align*}

Rustyn
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