Let-k-1-n-log-2-k-1994-find-the-positive-integer-n-

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 + \dots + {l o g}_{2} n = s

s = {l o g}_{2} (1 \times 2 \times 3 \times \dots \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 \dots

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

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 \dots \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

I n o w u n d e r s t a n d b e t t e r, t h a n k s s o m u c h .