Question

[Solved by @Kainui @thomas5267 @mathmath333 @ikram002p ] Find one prime divosor of
\[\large 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} + 23\]

Answer 1

Im going to go ahead and guess 23

Answer 2

Did you find any ?

Answer 3

yes

Answer 4

(i dont want to spoil the fun by saying anything about finding the factor :P)

Answer 5

I think I found one ahhahaha but I have no way of checking.

Answer 6

No wait, yes I do. Yeah I found a factor. I won't say it cause it's funny haha.

Answer 7

yes kainui got it xD

Answer 8

probably is the number itself, I bet it's a prime -.^

Answer 9

i would assume there is a factor x then
\(\large 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} + 23 \equiv 0 \mod x\)
\(\large 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} \equiv -23 \mod x\)

Answer 10

Just trial and error should work.

Answer 11

The number is small.

Answer 12

thomas also got it !

Answer 13

both have msged me the correct divosor and a legitimate method for proving it..

Answer 14

@ikram002p You are thinking in the right direction. Your answer inspired me to come up mine's.

Answer 15

i have several answers lol but idk away to make sure of it xD

Answer 16

and feeling stupid to post lol since already two of you knows :O

Answer 17

I found two factors other than 1 and the number itself lol =P

Answer 18

ive no idea how to even approach the question....that number looks bigger than everything that came into existence combined.......

Answer 19

:O

Answer 20

I can't find any others. One is hard enough for me...

Answer 21

lol i also thought one factor is good enough for me, but kainui seems to be having another factor... i don't understand his solution fully though il need to go through it few more time.. xD

Answer 22

well i thought of this xD

\(10^n+23 \)
for n which is factor of that power
hmm which also sounds does not work :O

Answer 23

@mathmath333 also got it!

Answer 24

I'll give a hint cause it's how I started. Try to write out the number. =P

Answer 25

xD

Answer 26

Should the stacked exponent read from top to bottom of left to right?

Answer 27

@ikram002p got it too

Answer 28

kainui, ikrram and mathmath claim to be having 2 divosors xD

Answer 29

ikram and mathmath have correct methods which i can understand easily for their 2 divisors !

Answer 30

I can't even write the number out. Too many zeros.

Answer 31

i bet that number doesnt exist in this universe with that many zeros !
ikram found 3 divosors!

Answer 32

Wow! I am thoroughly impressed.

Answer 33

last one was a joke :P

Answer 34

Haha well I guess I will reveal how I found my answers unless people are still wanting to try to figure it out.

Answer 35

ok reveal :O

Answer 36

i wanna see @thomas5267 solution though

Answer 37

\[
10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} + 23\equiv1^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}-1\equiv0\pmod3
\]

Answer 38

nice, any others ?

Answer 39

No I can't think of any other. 11 is not one of the prime divisor though.

Answer 40

Ok first one I found by trying to write it out, I got 100...0023 then I saw that if you add up all the digits you get 6, so it's divisible by 3.

Second one I just divided this number by 3. How did I do that? Well just like this number has 5 nines \[\Large 10^5-1=99999\] I noticed that we could rewrite the number as
\[\large 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}-1 + 24\] So really there are \[10^{10^{10^{10^{10^{10^{10^{10}}}}}}} \text{ nines}\] added to 24, so divide this by 3: \[(999...999 + 24)/3 = 333...333+8\] so Now there are the same number of 3s as there were nines, but if we add 8 we will get the last two digits change to 41 so really there are \[ \left( 10^{10^{10^{10^{10^{10^{10^{10}}}}}}}-2 \right) \text{ threes}\] followed by 41, so you could write it as: \[\Large 333...33341\] so that's the other number. Hahaha. =P

Answer 41

:O

Answer 42

brilliant xD thats another valid factor
so all in all we have found four divisors so far xD but only me knows all four factos haha!

Answer 43

Haha is the other one 13 and the number divided by 13?

Answer 44

thats @mathmath333 's divosor :)

Answer 45

3 and 13 give us another divosr : 39 xD

but yeah 39 is not so primy..

Answer 46

i told u that was a joke :P

Answer 47

I wonder if we can try to do some sort of trick to divide the number by 13 too.

Answer 48

can we find factors of
\[\large 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} +1\]

Answer 49

unthinkable

Answer 50

have some respect for large numbers ikram :P

Answer 51

Yeah seriously what's wrong with you @ikram002p ?

Answer 52

lol

Answer 53

I know a factor for this number where p is all primes greater than 23
\[\large 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} +p+1\]

Answer 54

BEAT THAT!

Answer 55

2 ?

Answer 56

2| (p+1)
2|10^n

so..

Answer 57

Yeah, but I also know 2 more numbers that are factors of that number... **cough**

Answer 58

1, 2, and the number itself. =P

Answer 59

i thought so.. leme think :)

Answer 60

lol okay

Answer 61

:P

Answer 62

4

Answer 63

for some p's

Answer 64

Wow.what the heck?

Answer 65

@FrownUpsideDown8767 - Welcome to Mathematics my friend :D

Answer 66

ok ok post some other question :D

Answer 67

im still working on proving 13 is a factor of the number in main question

Answer 68

so the question is not dead yet ikky :P

Answer 69

\(\large \begin{align} \color{black}{\dfrac{10^{10^n}+23}{13}\hspace{.33em}\\~\\
=\dfrac{((10^6) (10^4))^n+10}{13} \hspace{.33em}\\~\\
=\dfrac{((1)\cdot 100\cdot 100))^n-3}{13}\hspace{.33em}\\~\\
=\dfrac{((1)\cdot (81))^n-3}{13}\hspace{.33em}\\~\\
=\dfrac{3^{n~~ mod ~~3}-3}{13}\\ \normalsize {as~~ 3^3 mod ~~13=1} \hspace{.33em}\\~\\
=\dfrac{3^1-3}{13} \hspace{.33em}\\~\\
=0
}\end{align}\)

Answer 70

hahahaha @mathmath333 did it

Answer 71

wolfram

http://www.wolframalpha.com/input/?i=10%5E%2810%5E10%5E10%5E10%5E10%5E10%5E10%5E10%29+mod+13%3D

Answer 72

mathmath333 answer is so difficult to understand...

Answer 73

note that i used the cycle of mod 13 againts 3

Answer 74

I can't even understand your second step. Could you explain it?

Answer 75

the cycle of 3^\(\bf3\)=27 =1 mod 13)

so it is showing the cycle of 3

that why i applied mod 3 on \(n\)

Answer 76

this works in the cases for

\(\huge \begin{align} \color{black}{a^{b^c}\
}\end{align}\)

Answer 77

this stuff we do to find the remainders

Answer 78

mathmath's proof hinges on thae fact that we can write
\[\large 10^{10^{\cdots}} = 10^{10k} = (10^{10})^k\]
for some integer \(k\)

Answer 79

How does \(
\dfrac{10^{10^n}+23}{13}=\dfrac{((10^6) (10^4))^n+10}{13}
\)?

Answer 80

thats not true, below is true how ever
\[\dfrac{10^{10n}+23}{13}=\dfrac{((10^6) (10^4))^n+10}{13}\]

Answer 81

i dont see any flaw in rest of his work..

Answer 82

yep thats true

Answer 83

and the other factor is
769230769230......769230771
since 13|10023, 10000000023, 1000000000000023....
with quotients 771, 769230771, 769230769230771, ...

Answer 84

i will find if i can given a more elegant proof for this

Answer 85

thats a very interesting pattern @mathmate XD

Answer 86

10023/13=771
That's because 13|999999 and the quotient is 76923
and 10000000023-10023=9999990000+10023

Answer 87

Nice :) period of 1/13 is 6 so 13 | (10^6 - 1)

Answer 88

\(\dfrac{10^(6k+2)+23}{13}=\dfrac{10(6k)-1)*10000+10023}{13}=0~ mod~ 13\)

Answer 89

i have other solution though but seem way as mathmath thought off

Answer 90

Finally get it right:
\(\dfrac{10^{6k+2}+23}{13}=\dfrac{(10^{6k}-1)*10000+10023}{13}=0~ mod~ 13\)

Answer 91

10=-3 mod 13

(-3)^12=1 mod 13
(-3)^12n=1 mod 13
(-3)^12n+1=-3 mod 13
(-3)^12n+2=9 mod 13
(-3)^12n+3=-1 mod 13
(-3)^12n+4=3 mod 13
........


so we need to show 10^stuff is of the form 12n+4

Answer 92

@mathmath333 is a boss!!!

Answer 93

What happened here?
\[
\dfrac{((1)\cdot (81))^n-3}{13}
=\dfrac{3^{n\text{ mod }3}-3}{13}
\]

Answer 94

God I understand now. Definitely one of the hardest proof I have ever read.

Answer 95

i m trying to come up with another proof

Answer 96

I will try to make his proof more understandable.