Question

Find the last digit of 7 to the power 7 to the power 7 to the power 7 and so on upto 1 millionth term. Thanks in advance

Answer 1

Why:)

Answer 2

Hey estudies please help me dude
Its very urgent

Answer 3

what ive noticed is that if you look at 7 and its powers, 7^1 ends in a 7, 7^2 ends in a 9, 7^3 ends in a 3, 7^4 ends in a 1, 7^5 ends in a 7...and so it repeats all the time...mayb that wud help

Answer 4

there must be a patter yes? i
\[7^7\] ends in a 3

Answer 5

so i will be willing to bet they end up in a 3 from then on

Answer 6

Yes you are right but will you please explain step by step

Answer 7

sorry i am thinking...

Answer 8

ok lets try this explanation, although it may not be a very good one
taking powers of 7, the remainders (when divided by ten) aka the ones digits, repeat in the pattern 1,9,3,1,9,3 ...

Answer 9

nope i keep coming up with bad reasons still thinking

Answer 10

I can help you in this can i ?

Answer 11

please. i mean it seems obvious by i cannot seem to get a grip

Answer 12

There is a phenomena called cyclicity

Answer 13

This can be applied here

Answer 14

yes so it seems clear that if
\[7^7\] ends in a 3 then so must
\[7^{7^7}\] but i cannot come up with a good explanation.

Answer 15

7^3k and 7^7k periodicity for 3 ending....

Answer 16

oh yes now i can

Answer 17

This can be applied here

Answer 18

Thats good

Answer 19

ok
\[7^7\] ends in a 3 and therefore
\[7^{7^7}\] is
\[7^n\times 7^3\] where
\[n\equiv 1(mod 3)\]

Answer 20

no that is wrong

Answer 21

this problem is not as trivial as i thought or else my mind is shot

Answer 22

Oh no that is wrong dude

Answer 23

Yaaa it might happen because this is an olympiad question

Answer 24

I think it can't be 3^k, has to be 7^k by definition in the question.

Answer 25

yeah i see that it is wrong

Answer 26

7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401

7^5 = 16807
7^6 = 117649
7^7 = 823543


last digit of 7^(3k and 7k) = 3

set k = 7^m

Answer 27

OKKKKKKKKKKKK
Then ?

Answer 28

It's just induction.