Question

If we have (...(((7^7)^7)^7)^7...)^7 where 7 is raised to the 7 a thousand times, what is the last digit? Also, what is the second to last digit out of curiosity?

Answer 1

\[\LARGE 7^{7^{1000}}\] To be specific, you could rewrite it like this

Answer 2

hmmm how could rewrite it that way? just curious?

Answer 3

you*

Answer 4

Well that's just what it is. \[\LARGE (((7^7)^7)^7...)^7=7^{7*7*7*...*7}=7^{7^{1000}}\]

Answer 5

Oh i thought you were doing a different operation

i miss understood it :P

Answer 6

that number is extremely large hehe
did you mean that there is a way to get those last digit
and second to last

Answer 7

Yeah there definitely is a way to get the last digit and I don't know about the second to last, but probably.

Answer 8

sigh u are gonna turn into one of these mod freaks too now

Answer 9

1000/4 = perfect division so
its 1?

Answer 10

2nd last digit =0

Answer 11

wait for 7^1000 the 2nd last digit is 0 and 1 so

Answer 12

that means we will have 1 remainder with 4 so its 7!

Answer 13

7 is the last digit of 7^7^1000

Answer 14

7^1000 = ...........01
therefore
7^....01=
xx01/4= xx00/4 + 1/4 <--- remainder 1

Answer 15

I'm not sure how you got that answer but it's right haha.

Answer 16

01,7,49,343
^--- remainder 1

Answer 17

i did BS -.- pattern BS... now how do we prove it or do this with modulars

Answer 18

The way they showed it was that \[\LARGE 7=-1 \mod 4\] so in the exponent \[\LARGE (-1)^{1000}=1\] so then now we can say \[\LARGE 7^{7^{1000}}=7^1\] for the last digit.

But I don't really get how exactly this works lol

Answer 19

how do u think of mod

Answer 20

Basically if it's mod n then it's going to be some number from 0 to (n-1). I guess it's kind of interesting that they made it -1 but it's not completely wrong I mean that works I think since as long as your final answer lies inbetween it shouldn't matter I guess?

Answer 21

okay so like
7=3 mod 4
or 7=-1 mod 4