Question

Find the last two digits of 9^8^7. (By convention, exponent towers are evaluated from the top down, so 9^8^7 = 9^(8^7).)

Answer 1

You might want to look for patterns... Try seeing what the last two digits of 9^x are for a few values of x, and see if you can find a pattern.

Then use the pattern to help find the last two digits of 9^2097152 (which is 8^7). There might be a pattern in multiples of something, which you can then divide 2097152 by to shortcut.

Answer 2

oh. oohhh. i see the pattern. 1 then 9 then 1 then 9 forever and ever right? so then since 2097152 is even, then the units digit would be 9. but what about the tens digit?

Answer 3

wouldn't it be 1 in the units?

Answer 4

9^2 = 81

Answer 5

oh yeah. thanks @pgpilot326

Answer 6

(10 - 1)^ 8^7 use binomial expansion to get at last 2 terms?

Answer 7

We're interested in determining $$9^{8^7}\pmod 100$$

Answer 8

oops, that should be \(\pmod{100}\)!

Answer 9

mod 100 ?

Answer 10

http://www.wolframalpha.com/input/?i=9^%288^7%29

Last few digits:
081

Answer 11

wow @eliassaab, that is a really cool "calculator"

Answer 12

It is much much more than a calculator

Answer 13

yes, but how...

Answer 14

There is a pattern, but it's not all that simple unfortunately... and this is tedious.

Last two digits are the right column. ugh.

9^2 = 27
9^3 = 81
9^4 = 65 61
9^5 = 590 49
9^6 = 5314 41
9^7 = 47829 69
9^8 = 430467 21
9^9 = 3874204 89
9^10 = 34867844 01
9^11 = 313810596 09
9^12 = 2824295364 81
9^13 = 25418658283 29
9^14 = 228767924549 61
9^15 = 2058911320946 49
9^16 = 18530201888518 41
9^17 = 166771816996665 69
9^18 = 1500946352969991 21
9^19 = 13508517176729920 89
9^20 = 121576654590569288 01
9^21 = 1094189891315123592 09
9^22 = 9847709021836112328 81
9^23 = 88629381196525010959 29
9^24 = 797664430768725098633 61
9^25 = 7178979876918525887702 49
9^26 = 64610818892266732989322 41
9^27 = 581497370030400596903901 69

Answer 15

i didn't know what to call it :)

Answer 16

2, 8, 6, 4, 4, 6, 2, 8, 0, 0, 8, 2, 6, 4, 4, 6, 2, 8, 0, 0, 8, 2, 6, 4, 4, 6, 2, 8, 0, 0, 8, 2

blech. That doesn't make it very easy.

Answer 17

you could mod 10 the exponent and then take 9 to the remainder

90^10 = 1 mod 100 => 9^2 = last 2 digits, just as wolfram's calculated

Answer 18

thanks, everyone, for helping!

Answer 19

you need the euler totient function to find what power will mod to 1.

Answer 20

yay! thanks!

Answer 21

agh! i don't know who to give the medal to so?