Let Σ_(k = 1) ^n log_2 k = 1994 find the positive integer n ?


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Question Number 6811 by Tawakalitu. last updated on 28/Jul/16

$L e t \underset{k = 1}{\sum^{n}} {l o g}_{2} k = 1994$ $f i n d t h e p o s i t i v e i n t e g e r n ?$
Commented by Yozzii last updated on 29/Jul/16

$\underset{k = 1}{\sum^{n}} {l o g}_{2} k = {l o g}_{2} 1 + {l o g}_{2} 2 + {l o g}_{2} 3 + . . . + {l o g}_{2} n = s$ $s = {l o g}_{2} (1 \times 2 \times 3 \times . . . \times n) = {l o g}_{2} n!$ $\Rightarrow n! = 2^{s} . B u t s = 1994$ $\Rightarrow n! = 2^{1994} ?$
Commented by Tawakalitu. last updated on 29/Jul/16

$T h a n k s s o m u c h . . .$ $p l e a s e i s t h a t w h e r e i w i l l s t o p ?$
Commented by Yozzii last updated on 29/Jul/16
$If n=2⇒ n! has only the prime 2 in its prime factorisation. If n≥3, 3 must appear in the prime factorisation of n! with the power of 3 being at least one. n≥3⇒n!=1×2×3×{4×5×...×n} or 3 divides n!. This is for n∈N. In this problem, n∈N. We found that n!=2^(1994) ⋙6=3!⇒n is at least 3. But, if n∈N is at least 3 ⇒ 3∣n!. However, 3 does not divide 2^(1994) or 3∤n! while min(n)=3. So, no integer n exists satisfying n!=2^(1994) .$
$I f n = 2 \Rightarrow n! h a s o n l y t h e p r i m e 2 i n$ $i t s p r i m e f a c t o r i s a t i o n . I f n ⩾ 3, 3 m u s t$ $a p p e a r i n t h e p r i m e f a c t o r i s a t i o n o f$ $n! w i t h t h e p o w e r o f 3 b e i n g a t l e a s t o n e .$ $n ⩾ 3 \Rightarrow n! = 1 \times 2 \times 3 \times {4 \times 5 \times . . . \times n}$ $o r 3 d i v i d e s n! . T h i s i s f o r n \in N .$ $I n t h i s p r o b l e m, n \in N . W e f o u n d t h a t$ $n! = 2^{1994} ⋙ 6 = 3! \Rightarrow n i s a t l e a s t 3 . B u t,$ $i f n \in N i s a t l e a s t 3 \Rightarrow 3 ∣ n! . H o w e v e r,$ $3 d o e s n o t d i v i d e 2^{1994} o r 3 ∤ n! w h i l e$ $m i n (n) = 3 . S o, n o i n t e g e r n e x i s t s s a t i s f y i n g$ $n! = 2^{1994} .$
Commented by Tawakalitu. last updated on 29/Jul/16

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