Question

Find the last two digits of the perfect number
\[\large 2^{19936}\left(2^{19937}-1\right)\]

Answer 1

hmm i did such stuff i need some time to think
such stuff is in russians 8 grade which i'm studying now

Answer 2

bahaha

Answer 3

well 19936 and 19937 have difference of one i think this may be usful

Answer 4

u need to see the the last 2 digits of
2 to the 2 big numbers and
2 to the....36

Answer 5

Indeed thats a good observation! also 19937 is prime and this follows from the hypothesis that the number is a perfect number

Answer 6

it hought about trying to isolate 2^19366 as a sharable factor but it didn;t helped byffffff hmmm

Answer 7

idk if 19937 being a prime is useful in finidng the last two digits hmm

Answer 8

yeah

Answer 9

2 goes in
1
2
4
8
16
32
64
28
56
12
24
48
96

Answer 10

2^19936 * 2^1 =2^19937
i'm trying to isolate

Answer 11

I always forget. Does last two mean the one's and ten's digit or the other way around?

Answer 12

10s and 1s

Answer 13

yes ones and tens...

Answer 14

That is always confusing to me because I see them as the first digits of a number but I don't know anything. :p

Answer 15

123456

you always write 56 last :P

Answer 16

maybe distribute

Answer 17

not when I add numbers or subtract them :p

Answer 18

this looks hard.. im trying to find the units digit for now :/

Answer 19

2^39873 - 2^19936 if 2 is in power of 2 4 , 6 , 8, 10 and ........... the numebr will be also умут

Answer 20

hmm

Answer 21

will n mod 10 always give you the 1's digit of n?
and n mod 100 always give you 1's and 10's digit of n?

Answer 22

yes n mod 10 gives the units digit of n

Answer 23

i found theres a repetition in digits 2

Answer 24

there is a repetition

Answer 25

crre t

Answer 26

repitition in powers of 2 ?

Answer 27

yeah every 20 numbers

Answer 28

\[2^{20+x}\pmod{100} = 2^x \pmod{100}\]

?

Answer 29

12
24
48
96
92
84
68
36
72
44
88
76
52
04
08
16
32
64
28
56
12 <-repeat
24
.

Answer 30

yeah

Answer 31

every number repetas thata sequaence

Answer 32

thats an interesting pattern ! but how is it useful

Answer 33

>> mod((19936+19937),20)

ans =

13

>> mod((19936),20)

ans =

16

Answer 34

lets hope 3rrd digit is not needed

Answer 35

wee get 2 gihts and we need only 2 so this is useful

Answer 36

12 <----- 2^1 (techically 9 but...)
24
48
96
92
84
68
36
72
44 88 76 52 04 08 16 32 64 28 56 12 <-repeat

Answer 37

52 - 16

Answer 38

36

Answer 39

are the last 2 digits 36?

Answer 40

Very close!
answer is 56

Answer 41

56! it is in our secuane!!!

Answer 42

aye back to the drawing board

Answer 43

oh oops a mistake!

Answer 44

2^13= 8192
2^16=...36
soo
its 56

Answer 45

i forgot that thing was start at 2^9 anot not 2^1

Answer 46

http://prntscr.com/6dyroc

Answer 47

@dan815 yeah you are correct i thought it is much easier but i havent learned that stuff yer. we have been solving easier.

Answer 48

haha alexander are u in russia?

Answer 49

well no how not in russia but i'm studying suuing russian materials and books

Answer 50

wow! looks pretty neat.. how did you work out 13 and 16 ?

Answer 51

> mod((19936+19937),20)

ans =

13

>> mod((19936),20)

ans =

16

>> 2^13

Answer 52

i mean how do you know the last two digits of given number will be same as the last two digits of 2^13 - 2^16

Answer 53

so we check what 2^13 is and 2^16

Answer 54

oh because

Answer 55

u got 2^x(2^y-1)
so
2^(x+y)-2^x
=
.........92
-.......36
---------
you just hope the 2nd digit on the bottom wasnt higher than the top

Answer 56

and in our case it wasnt so no need for 3rd digit

Answer 57

92-36=56 so yeah

Answer 58

how did 19936+19937 became 13 ?

Answer 59

nvm i see XD

Answer 60

something to look into tho.... 36 was in the 20 number seequene and so was 56

Answer 61

is it possible that
2^n*(2^(n+1)-1) is always in the sequence of 20 digits

Answer 62

ah nvm thers a fail at 2^7(2^8-1)

Answer 63

the last two digits must be in the sequence of those 20 numbers whenever n+1 is prime

Answer 64

oh really lemme check some

Answer 65

wow thats cool

Answer 66

why is that happening?

Answer 67

actually it might not work always :/ one sec..

Answer 68

we know these :

1)
if \(2^{n-1}(2^n-1)\) is a perfect number then \(n\) is prime

2)
if \(2^n-1\) is a prime number then \(n\) is prime

Answer 69

lets see if those remainders really fall in the powers of 2 :

when \(n\) is prime, \(n\) will be in one of two forms : 4k+1, 4k+3

Answer 70

what is the def of perfect number

Answer 71

6 is perfect number because

6 = 1+2+3

a number which equals the sum of its proper divisors is called a perfect number

Answer 72

Claim 1 :

if \(n\) is in form \(4k+3\) then the last two digits of perfect number \(2^{n-1}(2^n-1)\) is always \(28\)

Answer 73

6 just happens to have first 3 natural numbers as its divosors

Answer 74

ah ya sry

Answer 75

but there is something related to what you found just now :

every perfect number is a triangular number

Answer 76

if
2^n-1 is a prime then n is prime hmm lemme think about this for a sec

Answer 77

kk

Answer 78

\(\large \color{black}{\begin{align} \normalsize \text{by euler totient function}\hspace{.33em}\\~\\
\phi{100}=40 \hspace{.33em}\\~\\
\dfrac{2^{19936}\left(2^{19937}-1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{2^{498\times 40+16}\left(2^{498\times 40+17}-1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{2^{16}\left(2^{17}-1\right)}{100}\hspace{.33em}\\~\\
\implies \dfrac{2^{16}\left(2^{17}-1\right)}{100}\hspace{.33em}\\~\\
\color{red}{\implies} \dfrac{2^{16}\left(2^{17}-1\right)}{100}\hspace{.33em}\\~\\

\end{align}}\)

this minimizes

Answer 79

that looks neat !!!

Answer 80

wow..

Answer 81

i dont know how that happened but wow..

Answer 82

euler was a cool guy

Answer 83

i m thinking that their can be much short method with binomial somehow

Answer 84

ya i was thinking about binomials too

Answer 85

there might be a simple method to find the 2nd digit with binomials expansions

Answer 86

yes binomial is very effective particularly in cases in \(\pmod {100}\) and \(\pmod {10}\)

Answer 87

oky i think i see it because if n is not prime then it will include a composition of primes other than 2

Answer 88

which cannot exist sinice its 2^p=2*n+2
and if n is a composition of 2s that is also not possible since ull have
2^P=2^k *(some product of primes)+2

Answer 89

wich simply cannt be since
u will have
2^P = 2*(2^(k-1)*(product of primes)/2 +1)
^ not possible of the product doesnt exist then its not possible because of the1

Answer 90

because that product of primes cannot include a 2

Answer 91

\(2^n - 1\) need not be a prime when \(n\) is prime

Answer 92

waiit whatt

Answer 93

\[2^n-1 \text{ is prime } \implies n \text{ is prime}\]

the converse wont hold

Answer 94

arghh i see okay

Answer 95

show me the proof for that statement 2)

Answer 96

ill check in a bit i got ppl overright now

Answer 97

which one

Answer 98

why
2^n-1 if its prime then n must be prime