Find-the-last-three-digits-of-2019-2019-

Question Number 84782 by mr W last updated on 16/Mar/20

F i n d t h e l a s t t h r e e d i g i t s o f 2019^{2019} .

Commented by jagoll last updated on 16/Mar/20

19^{2019} = {(18 + 1)}^{2019}

= \underset{i = 1}{\sum^{2019}} C_{i}^{2019} 18^{2019 - i} 1^{i}

that right ?

Commented by mr W last updated on 16/Mar/20

w h a t d o e s t h i s h e l p f o r f i n d i n g t h e

l a s t t h r e e d i g i t s o f 19^{2019}, s i r ?

Commented by mr W last updated on 16/Mar/20

i h a v e c h a n g e d t h e q u e s t i o n t o 2019^{2019},

s u c h t h a t i t i s a h u g e n u m b e r w h i c h

o n e {c a n}^{'} t “ {c a l c u l a t e}^{″} w i t h a c a l c u l a t o r .

Commented by jagoll last updated on 16/Mar/20

hahaha . . yes sir

Commented by jagoll last updated on 16/Mar/20

by using little fermat theorem ?

Commented by mr W last updated on 16/Mar/20

i {d o n}^{'} t k n o w t h i s t h e o r e m . i {d o n}^{'} t k n o w

i f t h e r e i s a s t a n d a r d m e t h o d t o s o l v e

t h i s k i n d o f q u e s t i o n s . b u t i t h i n k

i t i s p o s s i b l e t o s o l v e t h e m u s i n g

c o n v e n t i o n a l m e a n s .

Commented by Tony Lin last updated on 16/Mar/20

2019^{2019}

= {(2000 + 19)}^{2019}

= C_{0}^{2019} 19^{2019} + (C_{1}^{2019} 2000^{1} \times 19^{2018} + \cdot \cdot \cdot)

↑

10^{3} \times \cdot \cdot \cdot

19^{2019}

= {(20 - 1)}^{2019}

= C_{0}^{2019} {(- 1)}^{2019} + C_{1}^{2019} 20 \times {(- 1)}^{2018} +

C_{2}^{2019} 20^{2} \times {(- 1)}^{2017} + C_{3}^{2019} 20^{3} \times {(- 1)}^{2016}

+ \cdot \cdot \cdot

= 10956448923979 + 10^{3} \times \cdot \cdot \cdot

\to t h e l a s t t h r e e d i g i t s i s 979

Commented by mr W last updated on 16/Mar/20

{t h a t}^{'} s c o r r e c t s i r! {i t}^{'} s t h a t w h a t i m e a n t .

Answered by mr W last updated on 16/Mar/20

h e r e I^{'} l l s h o w y o u a g e n e r a l w a y I

d e v e l o p e d w h i c h a l w a y s w o r k s f o r

H o w t o f i n d t h e l a s t t h r e e d i g i t s o f

a h u g e n u m b e r l i k e 2019^{2019} ?

w e k n o w

{(1000 a + b)}^{n} = \underset{k = 0}{\sum^{n}} C_{k}^{n} {(1000 a)}^{n - k} b^{k}

= \underset{k = 0}{\sum^{n - 1}} C_{k}^{n} a^{n - k} b^{k} 1000^{n - k} + b^{n}

t h e f i r s t t e r m s \underset{k = 0}{\sum^{n - 1}} C_{k}^{n} a^{n - k} b^{k} 1000^{n - k} a r e

a m u l t i p l e o f 1000, i t s l a s t t h r e e d i g i t s

a r e 000, t h e r e f o r e t h e l a s t t h r e e d i g i t s

o f {(1000 a + b)}^{n} a r e t h e l a s t t h r e e d i g i t s

o f b^{n} . {l e t}^{'} s d e n o t e i t a s

{(1000 a + b)}^{n} \overset{3}{=} b^{n}

i t m e a n s t h e l a s t t h r e e d i g i t s f r o m

{(1000 a + b)}^{n} a r e e q u a l t o t h e l a s t t h r e e

d i g i t s f r o m b^{n} .

{i t}^{'} s o b v i o u s t h a t

c {(1000 a + b)}^{n} \overset{3}{=} {c b}^{n}

n o w {l e t}^{'} s l o o k a t 2019^{2019} .

2019^{2019} = {(2000 + 19)}^{2019}

\overset{3}{=} 19^{2019} = 19^{3} \times {(19^{4})}^{504} = 19^{3} \times {(130000 + 321)}^{504}

\overset{3}{=} 19^{3} \times 321^{504} = 19^{3} \times {(103000 + 41)}^{252}

\overset{3}{=} 19^{3} \times 41^{252} = 19^{3} \times {(2825000 + 761)}^{63}

\overset{3}{=} 19^{3} \times 761^{63} = 19^{3} \times 761 \times {(579000 + 121)}^{31}

\overset{3}{=} 19^{3} \times 761 \times 121^{31} = 19^{3} \times 761 \times 121 \times {(14000 + 641)}^{15}

\overset{3}{=} 19^{3} \times 761 \times 121 \times 641^{15} = 19^{3} \times 761 \times 121 \times 641 \times {(410000 + 881)}^{7}

\overset{3}{=} 19^{3} \times 761 \times 121 \times 641 \times 881^{7} = 19^{3} \times 761 \times 121 \times 641 \times 881 \times {(776000 + 161)}^{3}

\overset{3}{=} 19^{3} \times 761 \times 121 \times 641 \times 881 \times 161^{3} = 761 \times 121 \times 641 \times 881 \times {(3000 + 59)}^{3}

\overset{3}{=} 761 \times 121 \times 641 \times 881 \times 59^{3}

\overset{3}{=} 761 \times 121 \times 641 \times 881 \times 379

\overset{3}{=} 761 \times 121 \times 641 \times 899

\overset{3}{=} 761 \times 121 \times 259

\overset{3}{=} 761 \times 339

\overset{3}{=} 979

i . e . t h e l a s t t h r e e d i g i t s o f 2019^{2019}

a r e 979 .

Commented by mr W last updated on 16/Mar/20

a n o t h e r e x a m p l e :

17^{1789} = 17 \times {(17^{2})}^{894} = 17 \times {(200 + 89)}^{894}

\overset{2}{=} 17 \times 89^{894}

\overset{2}{=} 17 \times 21^{447} = 17 \times 21 \times {(21^{2})}^{223}

\overset{2}{=} 17 \times 21 \times 41^{223} = 17 \times 21 \times 41 \times {(41^{2})}^{111}

\overset{2}{=} 17 \times 21 \times 41 \times 81^{111} = 17 \times 21 \times 41 \times 81 \times {(81^{2})}^{55}

\overset{2}{=} 17 \times 21 \times 41 \times 81 \times 61^{55} = 17 \times 21 \times 41 \times 81 \times 61 \times {(61^{2})}^{27}

\overset{2}{=} 17 \times 21 \times 41 \times 81 \times 61 \times 21^{27} = 17 \times 41 \times 81 \times 61 \times {(21^{2})}^{14}

\overset{2}{=} 17 \times 41 \times 81 \times 61 \times 41^{14}

\overset{2}{=} 17 \times 41 \times 81 \times 61 \times 81^{7} = 17 \times 41 \times 61 \times 81^{8}

\overset{2}{=} 17 \times 41 \times 61 \times 61^{4}

\overset{2}{=} 17 \times 41 \times 61 \times 21^{2}

\overset{2}{=} 17 \times 41 \times 61 \times 41

\overset{2}{=} 17 \times 41 \times 1

\overset{2}{=} 97

i . e . t h e l a s t t w o d i g i t s o f 17^{1789} a r e 97 .

Commented by jagoll last updated on 16/Mar/20

amaZing

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