Find the last three digits of 2019^(2019) .


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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

	$amaZing$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum