How-many-digits-are-in-3-5000-

Question Number 6914 by Tawakalitu. last updated on 02/Aug/16

H o w m a n y d i g i t s a r e i n 3^{5000}

Commented by Rasheed Soomro last updated on 03/Aug/16

3^{1} = 3 (1 d i g i t), 3^{2} = 9 (1 d i g i t)

3^{3} = 27 (2 d i g i t s), 3^{4} = 81 (2 d i g i t s)

3^{5} = 243 (3 d i g i t s), 3^{6} = 729

3^{7} = 2187, 3^{8} = 6561 (4 d i g i t s)

⋮

L e t d i s n u m b e r o f d i g i t s i n l h s

W e c a n e a s i l y c o n c l u d e t h a t

3^{2 d - 1} (o d d p o w e r) a n d 3^{2 d} (e v e n p o w e r) b o t h h a v e d d i g i t s

N o w 3^{5000} = 3^{2 d}

2 d = 5000 \Rightarrow d = \frac{5000}{2} = 2500

3^{5000} = 3^{2 (2500)}

H e n c e n u m b e r o f d i g i t s = \underline{\overset{―}{∣ 2500 ∣}} i n 3^{5000}

Commented by Tawakalitu. last updated on 02/Aug/16

T h a n k s s o m u c h

Commented by sandy_suhendra last updated on 02/Aug/16

I t h i n k :

3^{2} h a s 1 d i g i t b e c a u s e \frac{2}{2} = 1

3^{4} h a s 2 d i g i t s b e c a u s e \frac{4}{2} = 2

3^{6} h a s 3 d i g i t s b e c a u s e \frac{6}{2} = 3

I f n = e v e n n u m b e r s o 3^{n} h a s \frac{n}{2} d i g i t s

3^{5000} h a s \frac{5000}{2} = 2500 d i g i t s

Commented by Rasheed Soomro last updated on 03/Aug/16

T h a n k s S a n d y . I h a d m a d e a m i s t a k e i n m y a n s w e r .

N o w I h a v e c o r r e c t e d i t .

Answered by nume1114 last updated on 03/Aug/16

I f n h a s 1 d i g i t, 0 ⩽ \log_{10} x < 1

n h a s 2 d i g i t s, 1 ⩽ \log_{10} n < 2

n h a s d d i g i t s, d - 1 ⩽ \log_{10} n < d

\log_{10} 3^{5000} = 5000 \times \log_{10} 3

\approx 5000 \times 0.47712

= 2385.6

2385 ⩽ \log_{10} 3^{5000} \approx 2385.6 < 2386

S o, 3^{5000} h a s 2386 d i g i t s ? ? ?

Commented by Yozzii last updated on 03/Aug/16

W o l f r a m s a y s t h i s i s c o r r e c t .

Commented by Rasheed Soomro last updated on 03/Aug/16

I f t h i s i s c o r r e c t t h e n w h a t i s

w r o n g i n m y l o g i c ? (S e e m y c o m m e n t a b o v e)

Commented by FilupSmith last updated on 03/Aug/16

is there a way to prove your logic ?

Commented by Yozzii last updated on 03/Aug/16

W o l f r a m i n d i c a t e s t h a t t h e p a t t e r n

y o u h a v e f o u n d w o r k s f o r s m a l l d \in N .

I t {d o e s n}^{'} t w o r k f o r d = 15 f o r e x a m p l e .

3^{29} h a s 14 d i g i t s w h i l e 3^{30} h a s 15 d i g i t s .

3^{5000} = 1 \times 10^{5000 {l o g}_{10} 3}

0.476 < {l o g}_{10} 3 < 0.48

\Rightarrow 10^{2380} = 10^{0.476 \times 5000} < 10^{5000 {l o g}_{10} 3} < 10^{5000 \times 0.48} = 1 \times 10^{2400}

\Rightarrow 3^{5000} h a s d e c i m a l l e n g t h b e t w e e n 2381 a n d 2401

T h i s i s b a s e d o n a p p r o x i m a t i o n s o f {l o g}_{10} 3 .

S o, a g o o d a p p r o x i m a t i o n i s

{l o g}_{10} 3 \approx 0.477121254

\Rightarrow 0.47712 < {l o g}_{10} 3 < 0.47720

\Rightarrow 10^{5000 \times 0.47712} < 3^{5000} < 10^{5000 \times 0.4772}

10^{2385.6} < 3^{5000} < 10^{2386}

10^{0.6} \times 10^{2385} < 3^{5000} < 10^{2386}

\Rightarrow 3^{5000} m u s t h a v e 2386 d i g i t s s i n c e i t c a n n o t

e x c e e d 1 \times 10^{2386} w h i c h h a s 2387 d i g i t s (1 \times 10^{2386} i s s m a l l e s t i n t e g e r > 0 w i t h 2387 d i g i t s)

b u t i s g r e a t e r t h a n 10^{0.6} \times 10^{2385} w h i c h h a s

2386 d i g i t s i f w e a p p r o x i m a t e 10^{0.6} t o a n

i n t e g e r (10^{0.6} \approx 4) .

Commented by nume1114 last updated on 03/Aug/16

Y o u s a i d 3^{2 d - 1} a n d 3^{2 d} b o t h h a v e d d i g i t s,

b u t a c t u a l l y i t i s i n c o l l e c t .

L e t D (n) b e n u m b e r o f d i s i t s i n n .

D (3^{2 d - 1}) = D (3^{2 d}) = d

D (3^{2 d + 1}) = D (3^{2 (d + 1) - 1}) = d + 1

D (3^{2 d}) \neq D (3^{2 d + 1})

B u t, w h e n d = 11,

3^{21} \approx 1.05 \times 10^{10} D (3^{21}) = 11

3^{22} \approx 3.14 \times 10^{10} D (3^{22}) = 11

3^{23} \approx 9.41 \times 10^{10} D (3^{23}) = 11 .

S o, D (3^{2 d}) \neq D (3^{2 d + 1}) i s n o t t r u e .

Commented by Rasheed Soomro last updated on 04/Aug/16

T H α n KS S s s t o a l l o f y o u!

T h i s i s a m a z i n g k n o w l e d g e f o r m e

t h a t t h e p a t t e r n {d o e s n}^{'} t w o r k f o r l a r g e r

n u m b e r s! T h a n k s a g a i n t o s h a r e t h i s .

I l e a r n t t h a t p a t t e r n s a r e n o t s o m u c h

r e l i a b l e f o r a l w a y s .

I s a i d t h a t

D (3^{2 d - 1}) = D (3^{2 d}) = d

B u t 2 d - 1 a n d 2 d a r e a n o d d n u m b e r

a n d a n e v e n n u m b e r n e x t t o i t .

w h e r e a s 2 d + 1 a n d 2 d a r e a n e v e n n u m b e r

a n d a n o d d n u m b e r n e x t t o i t .

M y p a t t e r n i s o n l y a b o u t a n e v e n n u m b e r

a n d i t s p r e v i o u s o d d n u m b e r .

(N o t a n e v e n n u m b e r a n d i t s n e x t

o d d n u m b e r f o r w h i c h y o u h a v e a r g u e d .)

N u m b e r : 3^{2 d - 1} 3^{2 d} ∣ 3^{2 d + 1} 3^{2 d + 2} ∣ \dots .

d i g i t s : d d ∣ d + 1 d + 1 ∣

P l e a s e m e n t i o n m e i f I a m w r o n g .

Commented by Tawakalitu. last updated on 03/Aug/16

T h a n k s s o m u c h . . i r e a l l y a p p r e c i a t e y o u r e f f o r t

Commented by nume1114 last updated on 04/Aug/16

S o r r y, m y e x p l a i n {w a s n}^{'} t g o o d .

B u t, 3^{24} \approx 2.82 \times 10^{11}, D (3^{24}) = 12 .

R e m e m b e r, D (3^{23}) = 11 .

S o, D (3^{2 d - 1}) \neq D (3^{2 d}) w h e n d = 12 .

Commented by Rasheed Soomro last updated on 04/Aug/16

T h a n k S! T h i s i s w h a t Y o z z i h a v e s a i d

t h a t f o r l a r g e r n u m b e r s t h e p a t t e r n {d o e s n}^{'} t

w o r k .